Help with Urgent Physics Incline Problem

[Stride]

Urgent help with physics incline problem!

A block of mass m is on an inclined ramp. The coefficient of static friction between the block and the ramp is µs. There is also a wall on the ramp situated below the block (see diagram).

http://img15.imageshack.us/img15/6056/image13pj.jpg

a) The angle of the ramp is increased. Up to a particular angle, θ, the force of static friction is able to hold the block in place. Find this angle.

b) For angles greater than θc, the wall needs to assist static friction in holding the block in place. Draw a free-body diagram to show this, and write the equations describing the forces acting on the block. Call the force from the wall, Fwall.

c) Solve the above equation for Fwall. Divide the equation you get for Fwall by the weight of the block to see what fraction of the weight is supported by the wall as a function of θ. Does your answer make sense? Explain why it does or does not.

 

[Stride]

heres what i have

b) Fnety= Fn-mg2cosθ
m1a= mg1sinθ-Usm1gcosθ-Fwall
m2a= mgsinθ-Usm2gcosθ

c) Fwall= mgsinθ-Usm1gcosθ


Fwall/m1g= (m1gsinθ-Usm1gcosθ)/m1g = Sinθ-Uscosθ

look good?

 

[astrokreng]

a) To find the angle θ, we can use the equation θ = tan^-1(µs), where µs is the coefficient of static friction. This angle represents the maximum angle at which the force of static friction is able to hold the block in place.

b) The free-body diagram for this situation would include the weight of the block acting downwards, the normal force from the ramp acting perpendicular to the surface, and the force of static friction acting parallel to the surface of the ramp. In addition, there would be a force from the wall, Fwall, acting in the opposite direction to the force of static friction. The equations describing the forces acting on the block would be:

∑Fy = N - mg = 0 (since the block is not moving in the y-direction)
∑Fx = Ff - Fwall = 0 (since the block is not moving in the x-direction)

c) Solving for Fwall, we get Fwall = µsN. Dividing this equation by the weight of the block (mg), we get:

Fwall/mg = µsN/mg

This can be simplified to:

Fwall/mg = µs(cosθ)

This equation shows that the fraction of the weight supported by the wall, Fwall/mg, is dependent on the angle of the ramp, θ, and the coefficient of static friction, µs.

This answer makes sense because as the angle of the ramp increases, the force of static friction must also increase in order to hold the block in place. This means that the wall must support a larger fraction of the weight of the block in order to assist static friction. As the angle of the ramp approaches the maximum angle θ, the force of static friction approaches its maximum value and the wall must support almost all of the weight of the block.