Download - Pse 5e Problem 5.61
A crate of weight Fg is pushed by a force P on a horizontal floor. (a) If the coefficient of static friction is
Μs and P is directed at an angle Θ below the horizontal, show that the minimum value of P that will move
the crate is given by Px = Μs Fg secHΘL H1 - Μs tanHΘLL-1
In[2158]:= sumForcesX@forces_D :=
Apply@Plus, Map@Function@elt, elt@@1DD * Cos@elt@@2DDDD, forcesDDsumForcesY@forces_D :=
Apply@Plus, Map@Function@elt, elt@@1DD * Sin@elt@@2DDDD, forcesDD
In[2161]:= forces = 88Fn, 90 °<, 8Ff, 180 °<, 8Fg, 270 °<, 8P, 360 ° - Θ<<;
In[2179]:= 8sumForcesX@forcesD � m * 0,
sumForcesY@forcesD � m * 0,
Ff � Fn * Μs,
Μs ¹ 0,
Fg ¹ 0,
Sin@ΘD ¹ 0, Cos@ΘD ¹ 0
<;
Reduce@%, 8P, Fn, Ff<D
Out[2180]= -Cos@ΘD + Μs Sin@ΘD ¹ 0 && P � -
Fg Μs
-Cos@ΘD + Μs Sin@ΘD&&
Fn � Fg + P Sin@ΘD && Ff � P Cos@ΘD && Fg Μs Cos@ΘD Sin@ΘD ¹ 0
Find the minimum value of Pthat can produce motion
when Μs = 0.400, Fg = 100 N, and Θ = 0°, 15.0°, 30.0°,
45.0°, and 60.0°.
In[2183]:= P � -
Fg Μs
-Cos@ΘD + Μs Sin@ΘD�. 8Μs ® 0.4, Fg ® 100, Θ ® 0<
Out[2183]= P � 40.
In[2186]:= P � -
Fg Μs
-Cos@ΘD + Μs Sin@ΘD�. 8Μs ® 0.4, Fg ® 100, Θ ® 15 °<
Out[2186]= P � 46.3823
In[2187]:= P � -
Fg Μs
-Cos@ΘD + Μs Sin@ΘD�. 8Μs ® 0.4, Fg ® 100, Θ ® 30 °<
Out[2187]= P � 60.0578
In[2188]:= P � -
Fg Μs
-Cos@ΘD + Μs Sin@ΘD�. 8Μs ® 0.4, Fg ® 100, Θ ® 45 °<
Out[2188]= P � 94.2809
In[2189]:= P � -
Fg Μs
-Cos@ΘD + Μs Sin@ΘD�. 8Μs ® 0.4, Fg ® 100, Θ ® 60 °<
Out[2189]= P � 260.434
2 PSE 5E PROBLEM 5.61.nb