Inclined Plane Force w/ Friction

AI Thread Summary
The discussion revolves around calculating the coefficient of static friction for a mass on an inclined plane with friction. A force of 59 Newtons is applied perpendicularly to the incline, and the incline angle at which the mass begins to slide is 56.7 degrees. Participants emphasize the importance of drawing a free body diagram (FBD) to accurately resolve gravitational forces into sine and cosine components. The normal force acting on the block is not equal to the applied force of 59 N, and calculations suggest that the coefficient of static friction is approximately 0.762, using the formula mg sin theta divided by the calculated normal force of 118 N. Understanding these forces is crucial for solving the problem accurately.
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Homework Statement


A mass of m = 11 kg is placed on incline plane with friction. A force with a magnitude of F = 59 Newtons is applied downward, perpendicular to the plane at all times. As the angle of the incline is increased from zero degrees it remains static until it reaches an angle of x = 56.7 degrees when it begins to slide. What is the coefficient of static friction for this situation?

I'm guessing that I sum up the forces as I would for a non-friction inclined plane problem, and then divide by the 59 N that they give you? Would I be assuming correctly?
 
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The normal reaction on the block due to the plane is not 59 N. Draw a complete FBD and you'll see. (Resolve the grav. force on the block into its sine and cosine components)
 
sArGe99 said:
The normal reaction on the block due to the plane is not 59 N. Draw a complete FBD and you'll see. (Resolve the grav. force on the block into its sine and cosine components)

I believe the answer is .762. mg sin theta / 118, where 118 is the calculated normal force?
 
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