You completed this assignment. Problem 5.76 Part A Block A in the figure (Figure

Question

You completed this assignment. Problem 5.76 Part A Block A in the figure (Figure 1) weighs 65.6 N. The coefficient of static friction between the block and the surface on which it rests is 0.29. The weight w is 10.6 N and the system is in equilibrium Find the friction force exerted on block A f= 65.5 Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining Figure 1 of 1 Part B Find the maximum weight for which the system will remain in equilibrium 45.0° a65.5 max Submit My Answers Give U Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

a) Assuming that there is a hypothetical "ring" in the middle that holds the 3 different strings in the figure, we can use the equilibrium equations of that "ring" because it should also be in equilibrium. Let the tension in the string connecting block A and the ring be T1, the one in the string connecting the ring and the hook be T2, and the one in the string connecting the ring and w be T3.

Fy = T2sin45 - T3 = 0
Fx = T2cos45 - T1 = 0

Since T3 = w = 10.6 N and we know that T2sin45 = T2cos45
T2sin45 = T3 = 10.6 N
T2cos45 = T2sin45 = 10.6 N
T2cos45 = T1 = 10.6 N

Now, since T1 = 10.6 N, we can calculate the equilibrium equations in block A. Note that T1 will be directed to the positive x direction because that's how tension in a string works. T1 will be counteracted by a friction force to the negative x direction and we opt to call this F. Summing the forces in the x direction,

Fx = T1 - F = 0
T1 = F = 10.6 N

b) Note that the relation between the weight w and the friction force is:
F = w
Computing for the maximum friction force that can be exerted by the ground on the block,
F = (coefficient of static friction x normal force)
In this case, the normal force is equal to the weight of the block,
F = (coefficient of static friction x weight of block A)
F = 0.29 x 65.6 N = 19.024 N (this is the maximum friction force that can be exerted by the ground to make the system remain in equilibrium)

F = w = 19.024 N

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