# Homework Help: Mechanics problem. Please clear my doubts regarding forces.

1. Aug 23, 2008

### tanzl

1. The problem statement, all variables and given/known data
In the ﬂgure below, the incline has mass M and is fastened to the stationary horizontal tabletop. The block of mass m is placed near the bottom of the incline and is released with a quick push that sets it sliding upward. It stops near the top of the incline, as shown in the ﬂgure, and then slides down again, always without friction. Find the force that the tabletop exerts on the incline throughout this motion.

http://i359.photobucket.com/albums/oo31/tanzl/PHYS1A_ProbSet2_08-09.jpg

2. Relevant equations

3. The attempt at a solution
My answer is (M+m)g since the resultant downward force should be the sum of the weight of incline plane and the weight of the mass. But, it looks too simple to be correct.
1)Although there is a normal force of magnitude mg*cos(theta) that will eventually balance the weight of the mass m since the mass is not moving downward o upward.
2)The initial push on the mass m is not necessary to be considered in the calculation because the force supplied is not constant and so the only force acting on mass m which parallel to the incline is mg*sin(theta).
3)Throughout the motion of mass, only two force were involved. The weight of mass m and the normal force of incline exerted on the mass.

2. Aug 23, 2008

### Shooting Star

You are neglecting the horizontal component of the normal reaction.
>
1)Although there is a normal force of magnitude mg*cos(theta) that will eventually balance the weight
>
The normal force N is equal to mgCosθ. I don’t quite understand what you mean by saying that it will eventually balance the weight.

The inclined wedge is in equilibrium due to three forces –the reaction R due to the tabletop, its own weight Mg, and the normal reaction N at the point of location of mass m. You know N in terms of m and θ. Draw a free body diagram and give it another try now. Resolve all the forces along the horizontal and vertical directions.

(Hint: R will not be in the vertical direction.)

3. Aug 24, 2008

### tanzl

This is my free body diagram for mass m.
http://i359.photobucket.com/albums/oo31/tanzl/freebodydiagram1.jpg
Fnormal = -Fg cos(theta)

This is my free body diagram for incline plane.
http://i359.photobucket.com/albums/oo31/tanzl/freebodydiagram2.jpg
I let Fx and Fy be the resolved vector of Ftable.
Fx=-Fg cos(theta) sin(theta) Fy=W-Fg [cos(theta)]^2
|Ftable^2|=|Fx|^2+|Fy|^2
By solving the equation, it yields

Ftable = g * sqrt {m^2 * [cos(theta)]^2 + M^2 + 2Mm [cos(theta)]^2}

4. Aug 24, 2008

### Shooting Star

Right. As a check, note that when θ=0, the reaction is simply (m+M)g, as is to be expected.

5. Aug 24, 2008

### tanzl

Thank you so much.