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
To solve this problem, we first need to resolve the applied force F into its horizontal and vertical components. The horizontal component Fx can be found by multiplying the magnitude of the force (250 N) by the cosine of the angle (20°):
Fx = F * cos(Θ) = 250 N * cos(20°) = 235 N
Similarly, the vertical component Fy can be found by multiplying the magnitude of the force by the sine of the angle:
Fy = F * sin(Θ) = 250 N * sin(20°) = 85.5 N
Next, to find the maximum static friction, we take the negative value of the horizontal component Fx:
Maximum static friction = -Fx = -235 N
To determine the normal force (Fn), we can use the fact that the sum of vertical forces is equal to zero when the object is in equilibrium. In this case, the vertical forces include the weight of the crate (mg) and the vertical component of the applied force (Fy). So, we can write:
Fn + Fy = mg
Since the weight of the crate is given as 55 kg, we can substitute the known values and solve for Fn:
Fn + 85.5 N = 55 kg * 9.8 m/s^2
Fn + 85.5 N = 539 N
Fn = 539 N - 85.5 N
Fn = 453.5 N
Finally, to find the coefficient of static friction (μs), we divide the maximum static friction by the normal force:
μs = Maximum static friction / Fn = (235 N / 453.5 N) = 0.52
So, the answers to the questions are:
(a) The normal force is 453.5 N
(b) The coefficient of static friction is 0.52