# Homework Help: Gravitational Acceleration

1. Dec 5, 2007

### amajorflaw

1. The problem statement, all variables and given/known data
a block of mass (m) is placed on a bigger triangular shaped block with a mass (M). The wedge is of angle (α) . The coefficient of static friction between the two blocks is μs . What must the force (F) be in order to keep the smaller block at the same height on the wedge-shaped block? The answer is an equation in terms of the coefficient of friction, M, m, the angle and the gravitational acceleration. A FBD is also required. All laws must be documented and used properly.

2. Relevant equations
Newton's 1st Condition
Gravitational Acceleration
Definition of Friction

3. The attempt at a solution

I have the answer, but I'm having trouble getting to it.
F = (M+m)((sinα-μscosα/cosα-μssinα))g

Last edited: Dec 5, 2007
2. Dec 5, 2007

You are missing another important theory, Newton's 2nd. Also, you should draw a free-body diagram to clearly communicate all of the forces being exerted on this body.

Casey

3. Dec 5, 2007

### rl.bhat

The acceleration of the system = a = F/(M + m )
the forces acting on the mass m are mg and ma. Resolve them into the components parallel and perpendicular to the surface of the wedge and find the condition for the equilibrium.

4. Dec 6, 2007

### amajorflaw

Yeah that's where I'm stuck. I'm not sure where all of the forces apply on the two blocks, so my free body diagram is wrong. It's not the equations that are troubling me. It's getting all of the parts to the FBD that is troubling me.

5. Dec 6, 2007

### rl.bhat

I'm not sure where all of the forces apply on the two blocks Both the blocks experiance equal force in thr horizontal direction. You have to draw FBD for the body of mass m. As I have already mensioned, resolve them into the components parallel and perpendicular to the surface of the wedge and find the expression for a.

6. Dec 6, 2007

### natugnaro

mgSin(alpha)-μsmgCos(alpha)=maCos(alpha)-μsmaSin(alpha)

forces acting on m in equilibrium.
forces due to gravity = forces due to F.

7. Dec 6, 2007

### rl.bhat

mgSin(alpha)-μsmgCos(alpha)=maCos(alpha)-μsmaSin(alpha)
That is correct. You can simplify this as g[Sin(alpha)-μsCos(alpha)]=a[Cos(alpha)-μsSin(alpha)]. Now substitute the value af a