Aim: How can we describe Newton’s 1st Law of Motion?

HW #4

Do Now:

Resolve the following vector into its horizontal and vertical components:

37°

200 N x-component

x = (200 N)cos37°

x = 200(4/5)

x = 160 N

y-component

y = (200 N)sin37°

y = 200(3/5)

y = 120 N

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Answer KeyHW 3

Newton’s 1st Law of Motion

Law of Inertia

If all the forces acting on an object are balanced:

a. an object at rest remains at rest

ΣFx = 0ΣFy = 0

b. an object in motion will remain in motion with a constant velocity

Inertia – the ability of an object to resist a change in motion

Objects at rest want to remain that way

Objects in motion want to remain that way

Inertia is dependent only on mass(the more mass, the more

inertia)

1. The sum of the forces on the object is zero in which of the cases?

(A)II only(B)III only(C)I and II only(D)I and III only(E)I, II, and III

Three objects can only move along a straight, level path. The graphs below show the position d of each of the objects plotted as a function of time t.

Graph I – constant velocity

Graph II – At rest

Graph III - Accelerating

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**1 minute**

2. Three forces act on an object. If the object is in translational equilibrium, which of the following must be true?

I. The vector sum of the three forces must equal zero.II. The magnitudes of the three forces must be equal.III. All three forces must be parallel.

(A)I only (B)II only(C)I and III only(D)II and III only(E)I, II, and III

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**1 minute**

II – 3 forces in the same direction all add up to a resultant force, yielding an unbalanced force

III – same reasoning as II

F

Fg

FN (also referred to as the apparent weight of the object in certain problems)

FF

For objects at rest or moving with a constant velocity:

F = Ff

FN = Fg

The key to the problems is the free-body diagram!

Fg = mg

Coefficient of Friction

Represented as µ

FF = µFN µs = coefficient of static friction(objects at rest)

µk = coefficient of kinetic friction(objects in motion)

µ is a unitless number between 0 (no friction) and 1 (100% friction)

Snow and ice have a low µ

Rubber has a high µ

FN

Fg = mg

F

FII

θ

θ

How can we solve for F║ and F┴

Mathematically, these θ’s are equal

Ff

For objects at rest or moving with a constant velocity on an incline:

Fll = Ff

FN = F┴

3. The mass of the block is most nearly(A) 1.0 kg(B) 1.2 kg(C) 1.6 kg(D) 2.0 kg(E) 2.5 kg

4. The magnitude of the normal force exerted on the block by the plane is most nearly(A) 10 N(B) 12 N(C) 16 N(D) 20 N(E) 33 N

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**2 minutes **

A plane 5 meters in length is inclined at an angle of 37°, as shown. A block of weight 20 newtons is placed at the top of the plane and allowed to slide down.

Fg = mg

20 N = m(10 m/s2)

m = 2 kg

FN = F┴

FN = FgcosθFN = (20 N)(4/5)FN = 16 N

F

θ

Fsinθ

Fcosθ

Resolve this into its components

5. A ball of mass m is suspended from two strings of unequal length as shown above. The tensions T1 and T2 in the strings must satisfy which of the following relations?

(A)Tl = T2 (B)T1 > T2 (C)T1 < T2(D)Tl + T2 = mg(E)T1‑T2 = mg

T2 has a larger vertical component and therefore is supporting more of the weight of mass m

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**1 minute**

6. A push broom of mass m is pushed across a rough horizontal floor by a force of magnitude T directed at angle as shown above. The coefficient of friction between the broom and the floor is . The frictional force on the broom has magnitude

(A) (mg + Tsin)

(B) (mg-Tsin)

(C) (mg+ Tcos)

(D) (mg-Tcos)

(E) mg

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**1 min 15 sec**

FN

mg Tsinθ

TcosθFF

FF = μFN

FF = μ(mg + Tsinθ)

ΣFy = 0

FN = mg + Tsinθ

7. A block of weight W is pulled along a horizontal surface at constant speed v by a force F, which acts at an angle of with the horizontal, as shown above. The normal force exerted on the block by the surface has magnitude

(A) W ‑ F cos(B) W‑Fsin(C) W(D) W + Fsin(E) W + Fcos

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**1 min 15 sec**

FN

W

Fsinθ

FcosθFF

ΣFy = 0

FN + Fsinθ – W = 0

FN + Fsinθ = W

FN = W - Fsinθ

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**1 min 15 sec**

FF = μFN

FF = μ(15 N + mg)

FF = 0.2(15 + 20)

FF = 0.2(35)

FF = 7 N

FN

15 N mg

ΣFy = 0

FN – 15 N – mg = 0

FN = 15 N + mg

Question #8

9. When an object of weight W is suspended from the center of a massless string as shown above. The tension at any point in the string is

(A)2Wcos

(B)½Wcos

(C) Wcos

(D) W/(2cos)

(E) W/cos

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**1 min 15 sec**

Tcosθ

W

Tcosθ

TsinθTsinθ

ΣFx = 0

Tsinθ – Tsinθ = 0

X – components cancel out

ΣFy = 0

Tcosθ + Tcosθ – W = 0

2 Tcosθ = W

T = W/(2 cosθ)

10. A uniform rope of weight 50 newtons hangs from a hook as shown above. A box of weight 100 newtons hangs from the rope. What is the tension in the rope?

(A)50 N throughout the rope

(B)75 N throughout the rope

(C)100 N throughout the rope

(D)150 N throughout the rope

(E)It varies from 100 N at the bottom of the rope to 150 N at the top.

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**1 min 15 sec**

100 N

T T

100 N 50 N

Bottom Top

ΣF = 0

T = 100 N

ΣF = 0

T = 100N + 50 N

T = 150 N

11. A box of uniform density weighing 100 newtons moves in a straight line with constant speed along a horizontal surface. The coefficient of sliding friction is 0.4 and a rope exerts a force F in the direction of motion as shown above.

a. On the diagram to the right, draw and identify all the forces on the box.

b. Calculate the force F exerted by the rope that keeps the box moving with constant speed.

FN

F

Fg

FF

ΣFx = 0

F = FF

F = μFN

F = μFg

F = (0.4)(100 N)

F = 40 N

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**2.5 minutes**

12. A 5.0‑kilogram monkey hangs initially at rest from two vines, A and B. as shown above. Each of the vines has length 10 meters and negligible mass.a. On the figure to the right, draw and label all of the forces acting on the monkey. (Do not resolve the forces into components, but do indicate their directions.)

b. Determine the tension in vine B while the monkey is at rest.

TATB

mg

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**5 minutes**