Suppose that a force F=250 N applied at 20° below the horizontal is just enough to start the motion of the 55 kg crate in Sample Problems 3.5 item 3. Find the (a) normal force and (b) coefficient of static friction.
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Here's Sample Problems 3 in the book I got the question from:
A 55 kg crate rests on a rough horizontal surface. A force of 250 N at an angle of 20° above the horizontal is enough to just start the motion of the crate. Find the (a) horizontal and vertical components of the applied force, (b) maximum static friction, (c) normal force, and (d) coefficient of static friction.
Given:
• m = 55 kg
• F = 250 N
• Theta = 20°
Solution:
The applied force F is resolved into its horizontal and vertical components as shown in the free-body diagram below.
a. Horizontal and vertical components of applied force:
Fx = (250 N) cos 20° = 235 N
Fy = (250 N) sin 20° = 85.5 N
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b. Maximum static friction:
-Fx = -235 N
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c. Normal force:
F N + 85.5 N = 539 N
F N = 453.5 N
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d. Coefficient of static friction:
Mu s = Maximum static friction/F N = 235 N / 453.5 N = 0.52
1 answer