Torque to linear acceleration.

[Yapper]

Urgent! Torque to linear acceleration. Please Help!

Homework Statement

Q1: A horizontal force F=1000N is applied on a 120kg fridge as shown below. If
friction is present, what is the magnitude of the fridge linear acceleration, and the
magnitude of the normal forces acting at A and B? (hint: this system is not in
static equilibrium, and any possible axis of rotation is obvious moving). .1 k .2 s friction coefficients the fridge only has contact with the floor at A and B.

[PLAIN]http://img714.imageshack.us/img714/3941/unledee.png

Homework Equations

Torque = r x F

The Attempt at a Solution

Torque from the horizontal force is 1200, torque from point A is .5(Fna), and torque from point B is in the opposite direction = .5(Fnb)

and Fnb + Fna = Fg = 120 * 9.8

I don't know how do figure this problem out at all, any help would be great.

 

[berkeman]

Homework Statement

Q1: A horizontal force F=1000N is applied on a 120kg fridge as shown below. If
friction is present, what is the magnitude of the fridge linear acceleration, and the
magnitude of the normal forces acting at A and B? (hint: this system is not in
static equilibrium, and any possible axis of rotation is obvious moving). .1 k .2 s friction coefficients the fridge only has contact with the floor at A and B.

[PLAIN]http://img714.imageshack.us/img714/3941/unledee.png

Homework Equations

Torque = r x F

The Attempt at a Solution

Torque from the horizontal force is 1200, torque from point A is .5(Fna), and torque from point B is in the opposite direction = .5(Fnb)

and Fnb + Fna = Fg = 120 * 9.8

I don't know how do figure this problem out at all, any help would be great.

Well, drawing a FBD of the fridge would be a good start. Remember the hint that the fridge is assumed to be in motion...

 

[Yapper]

I have a free body diagram. But i don't understand how to take that and figure out the linear acceleration and then find out the forces acting on A and B. I mean its not as simple as
1000 - .1(120kg)(9.8) = 120kg * a, a =7.35 m/s?

I need to know how to approach this problem after I have the FBD. how does the torque effect the translational? how do I figure out the differences in the normal forces at A and B since its not in equilibrium.

 

[berkeman]

I have a free body diagram. But i don't understand how to take that and figure out the linear acceleration and then find out the forces acting on A and B. I mean its not as simple as
1000 - .1(120kg)(9.8) = 120kg * a, a =7.35 m/s?

I need to know how to approach this problem after I have the FBD. how does the torque effect the translational? how do I figure out the differences in the normal forces at A and B since its not in equilibrium.

Can you please post your FBD and your equations for the balances of forces and torques?

 

[Yapper]

Fy = Fna + Fnb - Fg = 0

Fna + Fnb = Fg

Fx = F - fa - fb = F -.1(Fg)

Fx = 882.4 N ax = 7.35333 m/s^s

T = 1200 + .5Fna - .5Fnb + (fa x r) + (fb x r)

fa = .1Fna

fb = .1Fnb

[PLAIN]http://img198.imageshack.us/img198/3497/unled3t.png