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