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A box weighing 150 N is at rest on a horizontal floor. The coefficient of static friction is 0.4.

What is the smallest force F exerted eastward and upward at an angle of 350 with the horizontal that can start the box in motion?
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2 Answers

Hey Sierra -- we've got this nearly 3-4-5 /37 right triangle again ...
the eastward/upward force will distribute ~60% up and ~80% east ...
the up part will "lighten" the normal force on the floor: Fn = 150N less 0.6 up ...
F east must overcome 40% of Fn or F east = 60N minus 0.24 up ...
from approx. 3-4-5 tri -> F east is 145% of up => 1.45+0.25 up = 60N ...
up is 60/1.70 or 360/10 or 36N (3x12), making F east (4x12) => F min (5x12)
or F min ~60N ... Best wishes, ma'am :)   
Draw a free-body diagram with all four forces: weight, friction, normal force, and external force.
Let's call "eastward" the x-direction and "upward" the y-direction. The external force F has an x- and a y-component, given by F cos(35) and F sin(35), respectively. The x-component is canceled by the force of static friction, Fs, while the y-component, together with the normal force N, is canceled by the weight W of the box:
∑ Fx = F cos(35) - Fs =0
∑ Fy = F sin(35) + N - W =0
The second equation tells us that the normal force is
N = W - F sin(35)
We need the normal force because it is part of the force of friction:
Fs  = µ N = µ (W - F sin(35) )
Substitute this into the x-equation:
∑ Fx = F cos(35) - µ (W - F sin(35) ) =0
You could plug in your numbers at this point and solve for F. I prefer first solving for F algebraically before plugging in the values:
F ( cos(35) + µ sin (35)) = µ W
F = µ W /( cos(35) + µ sin (35))
F = 0.4 *150 /(cos (35) + 0.4 sin(35) ), which you can calculate yourself.