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?

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

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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, F

_{s}, while the y-component, together with the normal force N, is canceled by the weight W of the box:∑ F

_{x}= F cos(35) - F_{s}=0∑ F

_{y}= F sin(35) + N - W =0The 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:

F

_{s}= µ N = µ (W - F sin(35) )Substitute this into the x-equation:

∑ F

_{x}= F cos(35) - µ (W - F sin(35) ) =0You 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.