Equilibrium giving me some trouble

[mikeman19]

I just got done doing about 20 other problems but I can't figure this one out...

A uniform plank, with a length L of 7.63 m and a weight of 420 N, rests on the ground and against a frictionless roller at the top of a wall of height h = 5.21 m. The plank remains in equilibrium for any value of q= 70° or more, but slips if q < 70. Find the coefficient of static friction between the plank and the ground.

Any help would be appreciated

 

[Doc Al]

Start by identifying all the forces acting on the plank. Draw a diagram.

Then realize that the plank is in equilibrium, so:
∑F_x = 0
∑F_y = 0
∑Torques = 0 (about any point)

 

[M98Ranger]

!

I understand that equilibrium can be tricky, especially when dealing with different angles and forces. It's great that you have been practicing with other problems and I'm sure with some more practice, you will be able to figure this one out as well.

To find the coefficient of static friction, we can use the fact that the plank is in equilibrium. This means that the sum of all the forces acting on the plank must be equal to zero.

First, let's draw a free-body diagram of the plank. We have the weight of the plank acting downwards, and the normal force from the ground acting upwards. We also have a friction force acting parallel to the ground and in the opposite direction of the motion.

Since the plank is in equilibrium, we can set up the following equation:

ΣF = 0

This means that the sum of the forces in the x-direction and the sum of the forces in the y-direction must both equal zero.

In the x-direction, we have the friction force acting in the negative direction, and in the y-direction, we have the normal force acting in the positive direction.

ΣFx = -Ff = 0

ΣFy = N - mg = 0

Solving for the normal force, we get N = mg.

Now, we can use the fact that the plank will slip if q < 70. This means that the friction force must be equal to the maximum static friction force, which is μsN.

Substituting in our equation for N, we get the following:

-Ff = μsN = μsmg

Solving for μs, we get μs = -Ff/mg

Now, we can plug in the values given in the problem:

μs = -Ff/mg = -420/420(9.8) = -0.1

Since the coefficient of friction cannot be negative, we know that the maximum value for μs is 0.1.

I hope this explanation helps you understand how to approach this type of problem. Keep practicing and you will become more comfortable with equilibrium and other mechanics concepts. Good luck!