es 23 chapters

7/31/2019 ES 23 Chapters

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Moment of a ForceSubmitted by Romel Verterra on February 25, 2011 - 11:56pmMoment is the measure of the capacity or ability of the force to produce twisting or turningeffect about an axis. This axis is perpendicular to the plane containing the line of action of theforce. The magnitude of moment is equal to the product of the force and the perpendiculardistance from the axis to the line of action of the force. The intersection of the plane and theaxis is commonly called the moment center, and the perpendicular distance from the momentcenter to the line of action of the force is called moment arm.

From the figure above, O is the moment center and d is the moment arm. The moment M offorce F about point O is equal to the product of F and d.

226 - Moment of force about different pointsSubmitted by Romel Verterra on February 26, 2011 - 2:14pm

Problem 226InFig. P-226assuming clockwise moments as positive, compute the moment of force F =200 kg and force P = 165 kg about points A, B, C, and D.

Solution 226

Click here to show or hide the solution

Moment of force F about points A, B, C, and D:
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answer

answer

answer

answer

Moment of force P about points A, B, C, and D:

(this means that point A is on the line of action of force P) answer

answer

answer

You can also resolve P to horizontal and vertical components at point E then take themoment of these components at point C. The answer would be the same. Try it.

answer

227 - Moment of resultant force about a pointSubmitted by Romel Verterra on March 2, 2011 - 7:06amProblem 227Two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft a momentcenter O. Force P is 890 N directed up to the left at 30 with the horizontal and force Q is 445N directed up to the left at 60 with the horizontal. Determine the moment of the resultant ofthese two forces with respect to O.

Solution 227


(to the right)

(upward)

(counterclockwise) answer

The moment of resultant about O can be solved actually without the use of Rx and Ry. sum of the moment effect of each component is the same as the moment effect of the

components of resultant. Thus, . Try it. Also, the momabout point O can be found by solving the magnitude and finding the moment arm of thresultant then take the product of the two.

228 Intercepts of the resultant forceSubmitted by Romel Verterra on March 8, 2011 - 2:05pm

Problem 228Without computing the magnitude of the resultant, compute where the resultant of the fshown inFig. P-228intersects the x and y axes.
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Solution 228


to the right

upward

to the right

downward

clockwise

to the right

upward

x-intercept of the resultant

to the left of point O answer

y-intercept of the resultant

above point O answer

229 Y-coordinate of the point of application of the

forceSubmitted by Romel Verterra on March 22, 2011 - 1:59am

Problem 229InFig. P-229, find the y-coordinate of point A so that the 361-lb force will have a clockwmoment of 400 ft-lb about O. Also determine the X and Y intercepts of the line of actionforce.

Solution 229

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answer

Y-intercept of the line of action of force F

answer

X-intercept of the line of action of force F

answer

230 Distance from truss member to truss jointSubmitted by Romel Verterra on March 22, 2011 - 2:28am

Problem 230For the truss shown inFig. P-230, compute the perpendicular distance from E and from G tothe line BD. Hint: Imagine a force F directed along BD and compute its moment in terms of itscomponents about E and about G. Then equate these results to the definition of moment M =Fd to compute the required perpendicular distances.

Solution 230

Click here to show or hide the solutionLetd = length of member BDdx = 12 ftdy = 16 - 12 = 4 ft

Moment about point E

answer

Moment about point G
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answer

Checking (by Geometry):

(okay!)

(okay!)

231 Force P producing a clockwise moment about

originSubmitted by Romel Verterra on March 24, 2011 - 7:44am

Problem 231A force P passing through points A and B inFig. P-231has a clockwise moment of 300about O. Compute the value of P.

Solution 231

Click here to show or hide the solutionRatio and proportion

Moment at point O

down to the right from A to B answer
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232 Moment of a force about points O and BSubmitted by Romel Verterra on March 24, 2011 - 9:55pm

Problem 232InFig. P-231, the moment of a certain force F is 180 ftlb c lockwise about O and 90 ftlbcounterclockwise about B. If its moment about A is zero, determine the force.

Solution 232

Click here to show or hide the solutionMoment about O

Moment about B

Substitute xFy = 180 to the above equation

Thus, F = 75 lb downward to the right at x = 36.87 and x-intercept at (4, 0). ans

233 A force creating counterclockwise and clockw

momentsSubmitted by Romel Verterra on March 25, 2011 - 12:16am

Problem 233InFig. P-231, a force P intersects the X axis at 4 ft to the right of O. If its moment abou170 ftlb counterclockwise and its moment about B is 40 ftlb clockwise, determine its yintercept.
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Solution 233

Click here to show or hide the solutionResolve force P into components at its x-intercept

Resolve force P into components at its y-intercept

Thus, y intercept of force P is (0, -8/3). answer

CouplesSubmitted by Romel Verterra on April 14, 2011 - 10:03pm Couple is a system of forces whose magnitude of the resultant is zero and yet has amsum. Geometrically, couple is composed of two equal forces that are parallel to each otand acting in opposite direction. The magnitude of the couple is given by

Where are the two forces and is the moment arm, or the perpendicular distance bethe forces.

Couple is independent of the moment center, thus, the effect is unchanged in the followconditions.

The couple is rotated through any angle in its plane. The couple is shifted to any other position in its plane. The couple is shifted to a parallel plane.

In a case where a system is composed entirely of couples in the same plane or paralleplanes, the resultant is a couple whose magnitude is the algebraic sum of the originalcouples.

245 - Couple in the boxSubmitted by Romel Verterra on April 14, 2011 - 10:32pm
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Problem 245Refer to Fig. 2-24a. A couple consists of two vertical forces of 60 lb each. One force acts upthrough A and the other acts down through D. Transform the couple into an equivalent couplehaving horizontal forces acting through E and F.

Solution 245

Click here to show or hide the solutionanswer

246 - System with couples and forces on itSubmitted by Romel Verterra on April 14, 2011 - 11:04pm

Problem 246Determine the resultant moment about point A of the system of forces shown in Fig. P-Each square is 1 ft on a side.
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Solution 246


answer

247 - Resultant of couples in a 3-step pulleySubmitted by Romel Verterra on April 18, 2011 - 2:55pm

Problem 247

The three-step pulley shown inFig. P-247is subjected to the given couples. Compute tvalue of the resultant couple. Also determine the forces acting at the rim of the middle pthat are required to balance the given system.

Solution 247Click here to show or hide the solutionSolving for the resultant couple
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answer

Solving for equivalent couple acting in the middle pulley

The resultant couple is composed of two 63.3 lb forces. answer

248 - Broken handwheel replaced by lever to close a

gate valveSubmitted by Romel Verterra on April 18, 2011 - 3:29pm

Problem 248To close a gate valve it is necessary to exert two forces of 60 lb at opposite sides of ahandwheel 3 ft in diameter. Through an accident the wheel is broken and the valve must beclosed by a thrusting bar through a slot in the valve stem and exerting a force 4 ft out fromthe center. Determine the force required and draw a free-body diagram of the bar.

Solution 248

Click here to show or hide the solutionRequired couple to close the gate valve

Equivalent torque for the lever

answer

249 - Reactions at the bolts of speed reducer gear Submitted by Romel Verterra on April 18, 2011 - 3:49pm

Problem 249Fig. P-249represents the top view of a speed reducer which is geared for a four to onereduction in speed. The torque input at the horizontal shaft C is 100 lbft. The torque outhe horizontal shaft D, because of the speed reduction, is 400 lbft. Compute the torquereaction at the mounting bolts A and B holding the reducer to the floor. Hint: The torquereaction is caused by the unbalanced torque, which is a couple.
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Solution 249

Click here to show or hide the solutionThe couple resistance at the bolts is equivalent to the difference of torques on each shaft

R = 120 lb up at A and down at B. answer

250 Reaction at the supports of cantilever trussSubmitted by Romel Verterra on April 18, 2011 - 4:14pm

Problem 250The cantilever truss shown in Fig. P-250 carries a vertical load of 10.8 kN. The truss issupported by bearing at A and B which exert the forces Av, Ah, and Bh. The four forces constitute two couples which must have opposite moment effects to prevent movementtruss. Determine the magnitude of the supporting forces.

Solution 250


The given load and AV produce a counterclockwise couple
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Since Bh = Ah, the two are clockwise couple.

Thus,

Summary (answer)

251 Two forces producing a couple and a forceSubmitted by Romel Verterra on April 20, 2011 - 3:29pm

Problem 251A vertical force P at A and another vertical force F at B inFig. P-251produce a resultant of100 lb down at D and a counterclockwise couple C of 200 lbft. Find the magnitude anddirection of forces P and F.

Solution 251


answer

answer

252 Equivalent single force for a force and a couplSubmitted by Romel Verterra on April 20, 2011 - 3:55pm

Problem 252A force system consists of a clockwise couple of 480 Nm plus a 240 N force directed uthe right through the origin of X and Y axes at x = 30. Replace the given system by anequivalent single force and compute the intercepts its line of action with the X and Y ax

Solution 252

Click here to show or hide the solutionThe figure below shows the given force system
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The force system above can be replaced by the following three force system shown belowwhere the two pair into a couple which is the couple 480 Nm. The effect of these forceswould be the same as the original system.

The two opposing forces above will get canceled, thus, leaving a single force with the sameeffect as the three forces above which in turn has the same effect to the original system.

Thus, the equivalent single force is 240 N parallel to the given one and at a distance d from O. The intercepts of this single force are A(-4, 0) and B(0, 2.31). answer

253-254 Equivalent single force to replace a force a

a coupleSubmitted by Romel Verterra on April 20, 2011 - 4:06pm

Problem 253In Fig. P-253 a system of forces reduces to downward vertical force 400 lb through A pcounterclockwise couple of 800 lbft. Determine the single force that will produce an

equivalent effect.
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Click here to read or hide the solution to Problem 523

Problem 254

Rework Prob. 253 if the system reduces to a leftward horizontal force of 300 lb through pointA plus a clockwise couple of 750 lbft.

Click here to read or hide the solution to Problem 524

255 Equivalent loads to a compression member witheccentric loadSubmitted by Romel Verterra on April 21, 2011 - 10:04am

Problem 255A short compression member carries an eccentric load P = 200 lb situated 2 in. from the axis

of the member, as shown inFig. P-225. In strength of materials it is learned that the intstresses are determined from the equivalent axial load and couple into which P may beresolved. Determine the equivalent axial load and couple.

Solution 255


256 Twisting and bending effectsSubmitted by Romel Verterra on April 21, 2011 - 9:19pm

Problem 256A vertical shaft AB is 5 ft long and bolted to a rigid support at its lower end A. At its uppB is attached a horizontal bar BC which is 2 ft long. At the end of C is applied a force P
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lb. Force P is perpendicular to the plane containing points A, B, and C. Determine the twistingeffect of P on the shaft AB and the bending effect at point A.

Solution 256


Twisting effect

answer

Bending effect

answer

257 Horizontal couple equivalent to vertical forcesSubmitted by Romel Verterra on April 25, 2011 - 1:01pmProblem 257Replace the system of forces acting on the frame in Fig. P-257 by a resultant R at A and acouple acting horizontally through B and C.

Solution 257


Resultant of vertical forces

Moment about point A
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Location of R as measured from point A

Magnitude of horizontal couple at B and C

Thus the given system is equivalent to downward force of 50 kN at point A and clockwicouple of 220 kNm. The couple is represented by 110 kN horizontal forces at B and C.force at B is to the right and the force at C is to the left, producing the clockwisecouple. answer

Resultant of Concurrent Force SystemSubmitted by Romel Verterra on March 13, 2010 - 4:11pmResultant of a force system is a force or a couple that will have the same effect to the bboth in translation and rotation, if all the forces are removed and replaced by the resulta

The equation involving the resultant of force system are the following


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1.The x-component of the resultant is equal to the summation of forces in the x-direction.

2.The y-component of the resultant is equal to the summation of forces in the y-direction.

3.The z-component of the resultant is equal to the summation of forces in the z-direction.

Note that according to the type of force system, one or two or three of the equations abovewill be used in finding the resultant.

Resultant of Coplanar Concurrent Force SystemThe line of action of each forces in coplanar concurrent force system are on the same plane.

All of these forces meet at a common point, thus concurrent. In x-y plane, the resultant canbe found by the following formulas:

Resultant of Spatial Concurrent Force SystemSpatial concurrent forces (forces in 3-dimensional space) meet at a common point but dlie in a single plane. The resultant can be found as follows:

Direction Cosines


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Vector Notation of the Resultant

Where

011 Resultant of three forces acting in a ring

Submitted by Romel Verterra on September 3, 2010 - 6:47amProblem 011Three ropes are tied to a small metal ring. At the end of each rope three students are pulling,each trying to move the ring in their direction. If we look down from above, the forces anddirections they are applying are shown inFig. P-011. Find the net force on the ring due to thethree applied forces.

Solution 011

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Thus, the net force on the ring is 53.79 lb downward to the left at x = 17.16.

012 Resultant of two velocity vectorsSubmitted by Romel Verterra on September 4, 2010 - 12:48pm

Problem 012Find the resultant vector of vectors A and B shown inFig. P-012.

Solution 012: Component Method


The resultant vector R = 41.39 m/sec downward to the right at x = 72.70.

Another Solution: Vector Method


(okay!)
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downward to the right (ok!)

Another Solution: Geometry Method

Click here to show or hide the solutionCosine Law for the shaded triangle

(ok!)

By Sine Law

(okay!)

013 Resultant of three forces with angles greater than

90 degreeSubmitted by Romel Verterra on September 4, 2010 - 9:11pm

Problem 013Three vectors A, B, and C are shown in the figure below. Find one vector (magnitude anddirection) that will have the same effect as the three vectors shown inFig. P-013below.

Solution 013

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answer

014 Solving for force with given resultantSubmitted by Romel Verterra on September 8, 2010 - 5:25am

Problem 014From Fig. P-014, P is directed at an angle from x-axis and the200 N force is acting at a slope of 5 vertical to 12 horizontal.

a. Find P and if the resultant is 500 N to the right alongthe x-axis.

b. Find P and if the resultant is 500 N upward to the rightwith a slope of 3 horizontal to 4 vertical.

c. Find P and if the resultant is zero.

Solution 014

Click here to show or hide the solutionPart a: The resultant is 500N to the right along the x-axis

By Cosine law of the shaded triangle

answer

By Sine law

answer

Part b: The resultant is 500 N upward to the right with a slope of 3 horizontal to 4vertical
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answer

answer

Part c: The resultant is zero

The resultant is zero if P and the 200 N force areequal in magnitude, oppositely directed, andcollinear.

Thus, P = 200 N at = 157.38 answer

015 Solving for a force and its angle and angle of two

forces with given resultant

Submitted by Romel Verterra on September 9, 2010 - 5:39am

Problem 015Forces F, P, and T are concurrent and acting in the direction as shown in Fig. P-015.

a. Find the value of F and if T = 450 N, P = 250 N, = 30, and the resultant isN acting up along the y-axis.

b. Find the value of F and if T = 450 N, P = 250 N, = 30 and the resultant isc. Find the value of and if T = 450 N, P = 250 N, F = 350 N, and the resultant

zero.

Solution 015

Click here to show or hide the solutionPart a: Unknown force and direction with non-zero resultant

and
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answer

answer

Part b: Unknown force and direction with zero resultant

and

answer

answer

Part c: Unknown direction of two forces with zero resultant

and

Equation (1)

Equation (2)

Equation (1) + Equation (2)


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answer

From Equation (1)

answer

Resultant of Parallel Force SystemSubmitted by Romel Verterra on March 14, 2010 - 10:55pmCoplanar Parallel Force SystemParallel forces can be in the same or in opposite directions. The sign of the direction can be

chosen arbitrarily, meaning, taking one direction as positive makes the opposite directionnegative. The complete definition of the resultant is according to its magnitude, direction, andline of action.

Resultant of Distributed LoadsThe resultant of a distributed load is equal to the area of the load diagram. It is acting acentroid of that area as indicated. The figure below shows the three common distributeloads namely; rectangular load, triangular load, and trapezoidal load.

Rectangular Load

Triangular Load

Trapezoidal Load


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Spatial Parallel Force SystemThe resultant of parallel forces in space will act at the point where it will create equivalenttranslational and rotational (moment) effects in the system.

In vector notation, the resultant of forces are as follows...

Note:Two parallel forces that are equal in magnitude, opposite in direction, and not colinear willcreate a rotation effect. This type of pair is called a Couple. The placement of a couple in theplane is immaterial, meaning, its rotational effect to the body is not a function of its

placement. The magnitude of the couple is given by

Where F = the magnitude of the two equal opposing forces and d is the perpendiculardistance between these forces.

236 Computation of the resultant of parallel forces

acting on the leverSubmitted by Romel Verterra on March 26, 2011 - 12:57pmProblem 236

A parallel force system acts on the lever shown inFig. P-236. Determine the magnitudeposition of the resultant.

Solution 236


downward
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counterclockwise

to the right of A

Thus, R = 110 lb downward at 6 ft to the right of A. answer

237 Finding the resultant of parallel forces acting on

both sides of the rocker armSubmitted by Romel Verterra on March 26, 2011 - 1:26pm

Problem 237Determine the resultant of the four parallel forces acting on the rocker arm ofFig. P-237.

Solution 237


downward

clockwise

to the right of O

Thus, R = 50 lb downward at 4 ft to the right of point O. answer

238 Finding the resultant of trapezoidal loadingSubmitted by Romel Verterra on March 26, 2011 - 7:33pm

Problem 238The beam AB inFig. P-238supports a load which varies an intensity of 220 N/m to 890Calculate the magnitude and position of the resultant load.

S l i 238 S l i 239
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Solution 238


Thus, R = 3330 N downward at 3.6 m to the left of A. answer

239 Resultant of lift on the wing of an airplaneSubmitted by Romel Verterra on March 27, 2011 - 1:45am

Problem 239The 16-ft wing of an airplane is subjected to a lift which varies from zero at the tip to 360 lbper ft at the fuselage according to w = 90x1/2 lb per ft where x is measured from the tip.Compute the resultant and its location from the wing tip.

Solution 239


upward

th l t i i l t t th i ht f th i i l l t i th i ht f t i l
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Thus, R = 3840 lb upward at 9.6 ft from the tip of the wing. answer

Another Solution


okay!

okay!

240 How to locate the centroid of metal plate with

circular holeSubmitted by Romel Verterra on March 28, 2011 - 1:57am

Problem 240The shaded area inFig P-240represents a steel plate of uniform thickness. A hole of 4-in.diameter has been cut in the plate. Locate the center of gravity the plate. Hint:The weight of

the plate is equivalent to the weight of the original plate minus the weight of material cuaway. Represent the original plate weight of plate by a downward force acting at the cethe 10 14 in. rectangle. Represent the weight of the material cut away by an upward facting at the center of the circle. Locate the position of the resultant of these two forcesrespect to the left edge and bottom of the plate.

Solution 240

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Thus, the centroid is located at 6.8 in. to the right of left edge and 4.9 in. above the bottomedge. answer

241 Finding the resultant of vertical forces acting onthe Fink trussSubmitted by Romel Verterra on March 29, 2011 - 10:16pm

Problem 241Locate the amount and position of the resultant of the loads acting on the Fink truss inFig. P-241.

Solution 241

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Magnitude of resultant

downward

Location of resultant

to the right of A

Thus, R = 15 130 N downward at 3.62 m to the right of left support. answer

242 Finding the unknown two forces with given

resultantSubmitted by Romel Verterra on March 30, 2011 - 12:07pm

Problem 242Find the value of P and F so that the four forces shown in Fig. P-242 produce an upwaresultant of 300 lb acting at 4 ft from the left end of the bar.

Solution 242

Click here to show or hide the solutionSum of vertical forces

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answer

answer

243 Finding the magnitude and position of the

missing forceSubmitted by Romel Verterra on March 30, 2011 - 12:31pm

Problem 243The resultant of three parallel loads (one is missing in Fig. P-243) is 13.6 kg acting up at 3 mto the right of A. Compute the magnitude and position of the missing load.

Solution 243

Click here to show or hide the solutionSum of vertical forces

downward


Thus, F = 31.4 kg downward at 2.48 m to the right of left support. answer

Resultant of Non-Concurrent Force SystemSubmitted by Romel Verterra on May 11, 2011 - 3:55pmThe resultant of non-concurrent force system is defined according to magnitude, inclina

and position.

The magnitude of the resultant can be found as follows

The inclination from the horizontal is defined by

The position of the resultant can be determined according to the principle of moments.
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Where,Fx = component of forces in the x-direction

Fy = component of forces in the y-directionRx = component of thew resultant in x-directionRy = component of thew resultant in y-directionR = magnitude of the resultantx = angle made by a force from the x-axisMO = moment of forces about any point Od = moment armMR = moment at a point due to resultant forceix = x-intercept of the resultant Riy = y-intercept of the resultant R

Problem 260 - 261 | Resultant of Non-Concurrent

Force SystemSubmitted by Romel Verterra on May 11, 2011 - 5:12pm

Problem 260The effect of a certain non-concurrent force system is defined by the following data: Fx =+90 kN,Fy = -60 kN, and MO = 360 kNm counterclockwise. Determine the point at whichthe resultant intersects the x-axis.

Solution 260


The x-intercept is at 6 m to the left of the origin. answer

Problem 261In a certain non-concurrent force system it is found that Fx = -80 lb, Fy = +160 lb,and MO = 480 lbft in a counterclockwise sense. Determine the point at which the resuintersects the y-axis.

Solution 261


The y-intercept of the resultant is 6 ft above theorigin. answer

Problem 262 | Resultant of Non-Concurrent Force SystemSubmitted by Romel Verterra on May 11, 2011 - 5:50pm

Problem 262Determine completely the resultant of the forces acting on the step pulley shown inFig262.
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Solution 262


Thus, R = 1254.89 lb downward to the right at x = 44.21 and passes through the axle

Problem 263 | Resultant of Non-Concurrent Force

SystemSubmitted by Romel Verterra on May 11, 2011 - 6:23pm

Problem 263Determine the resultant of the force system shown inFig. P-263and its x and y interce

Solution 263

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Thus, R = 161.314 lb upward to the right at x = 21.69 and intercepts at (1.668, 0) and0.671).



Problem 264Completely determine the resultant with respect to point O of the force system shown inP-264.


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Solution 264

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Thus, R = 544.68 N upward to the right at x = 28.25. The intercepts of R are (-4.57, 0) and(0, 2.46).



Problem 265Compute the resultant of the three forces shown in Fig. P-265. Locate its intersection with Xand Y axes.

Solution 265


Solution 266
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Thus, R = 957.97 lb downward to the right at x = 32.19. The x-intercept is at 2.90 ft to theright of O and the y-intercept is 1.83 ft above point O.

Problem 266 | Resultant of Non-Concurrent ForceSystemSubmitted by Romel Verterra on May 12, 2011 - 2:18pm

Problem 266Determine the resultant of the three forces acting on thedamshown inFig. P-266and locateits intersection with the base AB. For good design, this intersection should occur within themiddle third of the base. Does it?


Righting moment

Overturning moment

Moment at the toe (downstream side - point B)
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Location of Ry as measured from the toe

(within the middle third)

Thus, R = 27 424.02 lb downward to the right at x = 79.91 and passes through the base at8.44 ft to the left of B which is within the middle third.



Problem 267The Howe roof truss shown inFig. P-267carries the given loads. The wind loads areperpendicular to the inclined members. Determine the magnitude of the resultant, itsinclination with the horizontal, and where it intersects AB.

Solution 267

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Thus, R = 10 778.16 N downward to the right at x = 68.2 passing 4.46 m to the right o



Problem 268The resultant of four forces, of which three are shown inFig. P-268, is a couple of 480

clockwise in sense. If each square is 1 ft on a side, determine the fourth force complete

Solution 268

Click here to show or hide the solutionLet F4 = the fourth force and for couple resultant, R is zero.

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Thus,

Assuming F4 is above point O

d is positive, thus, the assumption is correct that F4 is above point O.

Therefore, the fourth force is 200 lb acting horizontally to the left at 5.8 ft above pointO. answer


Problem 269RepeatProb. 268is the resultant is 390 lb directed down to the right at a slope of 5 to 1passing through point A. Also determine the x and y intercepts of the missing force F.

Solution 269

Click here to show or hide the solutionLet F4 = the fourth force
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Resolve F4 into components at the x-axis

Resolve F4 into components at the y-axis

Thus, F4 = 219.32 lb downward to the right at x = 43.15 with x-intercept ix = 3.27 to thof O, and y-intercept iy = 3.06 ft above point O.

Problem 270 | Resultant of Non-Concurrent ForceSystemSubmitted by Romel Verterra on May 12, 2011 - 5:40pm

Problem 270The three forces shown in Fig. P-270 are required to cause a horizontal resultant acting

through point A. If F = 316 lb, determine the values of P and T. Hint: Apply MR = MB todetermine R, then MR = MC to find P, and finally MR = MD or Ry = Y to compute T.


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Solution 270

Click here to show or hide the solutionFor horizontal resultant, Ry = 0 and Rx = R

answer

answer



Problem 271The three forces in Fig. P-270 create a vertical resultant acting through point A. If T is kto be 361 lb, compute the values of F and P.
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Solution 271

Click here to show or hide the solutionFor vertical resultant, Rx = 0 and Ry = R

answer

answer

Equilibrium of Force SystemSubmitted by Romel Verterra on July 13, 2011 - 6:31amThe body is said to be in equilibrium if the resultant of all forces acting on it is zero. Thetwo major types of static equilibrium, namely, translational equilibrium and rotationalequilibrium.

FormulasConcurrent force system
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Parallel Force System

Non-Concurrent Non-Parallel Force System

Problem 308 | Equilibrium of Concurrent Force


Problem 308The cable and boom shown inFig. P-308support a load of 600 lb. Determine the tensile

force T in the cable and the compressive for C in the boom.

Solution 308


answer

answer

Another Solution (By Rotation of Axes)



System
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(okay!)

(okay!)

Another Solution (By Force Polygon)


(okay!)

(okay!)


Problem 309A cylinder weighing 400 lb is held against a smooth incline by means of the weightless AB inFig. P-309. Determine the forces P and N exerted on the cylinder by the rod and tincline.

Solution 309

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answer

answer

Another Solution (By Rotation of Axes)


(ok!)

(ok!)

Another Solution (By Force Polygon)


answer
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(ok!)

(ok!)

Problem 310 - 311 | Equilibrium of Concurrent Force


Problem 310A 300-lb box is held at rest on a smooth plane by a force P inclined at an angle with theplane as shown inFig. P-310. If = 45, determine the value of P and the normal pressure Nexerted by the plane.

Solution 310


answer

Problem 311If the value of P in Fig. P-310 is 180 lb, determine the angle at which it must be inclinethe smooth plane to hold 300-lb box in equilibrium.

Solution 311


answer


SystemSubmitted by Romel Verterra on May 25, 2011 - 7:13am

Problem 312Determine the magnitude of P and F necessary to keep the concurrent force system in P-312 in equilibrium.

Determine the value of P and E to maintain equilibrium of the forces.
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Solution 312


answer

answer



Problem 313Figure P-313 represents the concurrent force system acting at a joint of a bridge truss.

Solution 313


answer
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answer

answer



Problem 314The five forces shown in Fig. P-314 are in equilibrium. Compute the values of P and F.

Solution 314


answer

Problem 315 | Equilibrium of Concurrent ForceSystemSubmitted by Romel Verterra on May 25, 2011 - 7:55am

Problem 315The 300-lb force and the 400-lb force shown in Fig. P-315 are to be held in equilibrium third force F acting at an unknown angle with the horizontal. Determine the values of and .

The correct position of F would be as shown below.
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Solution 315

Click here to show or hide the solutionBy Cosine Law

answer

answer



Problem 316Determine the values of and so that the forces shown in Fig. P-316 will be in equilib

Solution 316


By Cosine Law

answer
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answer



Problem 317The system of knotted cords shown inFig. P-317support the indicated weights. Compute thetensile force in each cord.

Solution 317

Click here to show or hide the solutionFrom the knot where 400-lb load is hanging

answer

answer
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From the knot where 300-lb load is hanging

answer

answer


SystemSubmitted by Romel Verterra on May 27, 2011 - 9:19pmProblem 318Three bars, hinged at A and D and pinned at B and C as shown inFig. P-318, form a four-linkmechanism. Determine the value of P that will prevent motion.

Solution 318

Click here to show or hide the solutionAt joint B

At joint C

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answer



Problem 319Cords are loop around a small spacer separating two cylinders each weighing 400 lb andpass, as shown inFig. P-319over a frictionless pulleys to weights of 200 lb and 400 lb .Determine the angle and the normal pressure N between the cylinders and the smoothhorizontal surface.

Solution 319

answer

answer

Problem 322 | Equilibrium of Force SystemSubmitted by Romel Verterra on May 31, 2011 - 9:57pm

Problem 322The Fink truss shown inFig. P-322is supported by a roller at A and a hinge at B. The gloads are normal to the inclined member. Determine the reactions at A and B. Hint: Repthe loads by their resultant.

Solution 322


answer
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Thus,RB = 4618.80 lb at 30 with horizontal answer

Another Solution


Problem 323The truss shown inFig. P-323is supported by a hinge at A and a roller at B. A load of 2is applied at C. Determine the reactions at A and B.
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(okay!)

From the Force Polygon

(okay!)

(okay!)

Problem 323 | Equilibrium of Force SystemSubmitted by Romel Verterra on May 31, 2011 - 10:40pm

Solution 323


answer

Thus,
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RA = 21.06 kN down to the left at 34.7 with the horizontal. answer

Another Solution


(okay!)

answer
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(okay!)

(okay!)

Problem 324 | Equilibrium of Three-force SystemSubmitted by Romel Verterra on June 2, 2011 - 11:23pm

Problem 324A wheel of 10-in radius carries a load of 1000 lb, as shown inFig. P-324. (a) Determine thehorizontal force P applied at the center which is necessary to start the wheel over a 5-in.block. Also find the reaction at the block. (b) If the force P may be inclined at any angle withthe horizontal, determine the minimum value of P to start the wheel over the block; the angleP makes with the horizontal; and the reaction at the block.

Solution 324Click here to show or hide the solutionPart (a)

answer

Part (b)

answer
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answer

answer

answer

Problem 325 | Equilibrium of Three-force SystemSubmitted by Romel Verterra on June 2, 2011 - 11:58pm

Problem 325Determine the amount and direction of the smallest force P required to start the wheel inFig.P-325over the block. What is the reaction at the block?

Solution 325

Click here to show or hide thesolution

answer
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answer

answer

answer

Problem 326 | Equilibrium of Force SystemSubmitted by Romel Verterra on June 3, 2011 - 11:07am

Problem 326The cylinders inFig. P-326have the indicated weights and dimensions. Assuming smoothcontact surfaces, determine the reactions at A, B, C, and D on the cylinders.

Solution 326


From the FBD of 200 kN cylinder

answer

answer
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answer

answer

From the FBD of 400 kN cylinder

answer

Problem 327 | Equilibrium of Force SystemSubmitted by Romel Verterra on June 4, 2011 - 1:02pmProblem 327Forces P and F acting along the bars shown inFig. P-327maintain equilibrium of pin ADetermine the values of P and F.

Solution 327


Equation (1)
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Equation (1)

Substitute F of Equation (1)

answer

From Equation (1)

answer

Problem 328 | Equilibrium of Force SystemSubmitted by Romel Verterra on June 4, 2011 - 1:14pm

Problem 328Two weightless bars pinned together as shown in Fig. P-328 support a load of 35 kN.Determine the forces P and F acting respectively along bars AB and AC that maintainequilibrium of pin A.

Solution 328


answer

answer

Equilibrium of Parallel Force SystemSubmitted by Romel Verterra on June 25, 2011 - 11:18amC di i f E ilib i f P ll l F
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Conditions for Equilibrium of Parallel ForcesThe sum of all the forces is zero.

The sum of moment at any point O is zero.

Problem 329 | Equilibrium of Force SystemSubmitted by Romel Verterra on June 4, 2011 - 1:29pm

Problem 329Two cylinders A and B, weighing 100 lb and 200 lb respectively, are connected by a rigid rodcurved parallel to the smooth cylindrical surface shown inFig. P-329. Determine theangles and that define the position of equilibrium.

Solution 329


From the figure

Thus,

answer

answer

Problem 332 | Equilibrium of Parallel Force System

Submitted by Romel Verterra on June 6, 2011 - 10:21pm

Problem 332Determine the reactions for the beam shown in Fig. P-332.

answer

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Solution 332


answer

Problem 333 | Equilibrium of Parallel Force SystemSubmitted by Romel Verterra on June 6, 2011 - 10:34pm

Problem 333Determine the reactions R1 and R2 of the beam in Fig. P-333 loaded with a concentrateof 1600 lb and a load varying from zero to an intensity of 400 lb per ft.

Solution 333


Problem 334Determine the reactions for the beam loaded as shown in Fig. P-334.
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answer

answer


Solution 334


answer

answer

http://www.mathalino.com/users/romel-verterrahttp://www.mathalino.com/reviewer/engineering-mechanics/problem-334-equilibrium-parallel-force-systemhttp://www.mathalino.com/reviewer/engineering-mechanics/problem-334-equilibrium-parallel-force-systemhttp://www.mathalino.com/reviewer/engineering-mechanics/problem-334-equilibrium-parallel-force-systemhttp://www.mathalino.com/users/romel-verterrahttp://www.mathalino.com/users/romel-verterrahttp://www.mathalino.com/users/romel-verterra


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Problem 335The roof truss in Fig. P-335 is supported by a roller at A and a hinge at B. Find the values ofthe reactions.

Solution 335


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Submitted by Romel Verterra on June 6, 2011 - 11:05pmProblem 336The cantilever beam shown in Fig. P-336 is built into a wall 2 ft thick so that it rests agapoints A and B. The beam is 12 ft long and weighs 100 lb per ft.

Solution 336

http://www.mathalino.com/reviewer/engineering-mechanics/problem-335-equilibrium-parallel-force-systemhttp://www.mathalino.com/reviewer/engineering-mechanics/problem-335-equilibrium-parallel-force-systemhttp://www.mathalino.com/users/romel-verterrahttp://www.mathalino.com/users/romel-verterrahttp://www.mathalino.com/reviewer/engineering-mechanics/problem-335-equilibrium-parallel-force-systemhttp://www.mathalino.com/reviewer/engineering-mechanics/problem-336-equilibrium-parallel-force-systemhttp://www.mathalino.com/reviewer/engineering-mechanics/problem-336-equilibrium-parallel-force-systemhttp://www.mathalino.com/reviewer/engineering-mechanics/problem-336-equilibrium-parallel-force-system


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Problem 337The upper beam in Fig. P-337 is supported at D and a roller at C which separates the upperand lower beams. Determine the values of the reactions at A, B, C, and D. Neglect the weightof the beams.

Solution 337


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