What is the acceleration of each block?

[siimplyabi]

1. The
problem statement, all variables and given/known data
A pair of blocks are connected by a massless rope and placed on a frictionless surface. The diagram shows the situation at the instant the blocks are released. The left side has a mass of 35 kg and an incline of 25 degrees to the horizontal. The right side has a mass of 20 kg and an incline of 40 degrees to the horizontal.

Homework Equations

[/B]

The Attempt at a Solution

My attempt was solving each blocks acceleration by doing
Σf(+x)= Fgsin(theta)=ma
fg = mg there fore mgsin(theta)=ma
the masses cancel out leaving
gsin(theta)=a
so acc of block 1 =
(9.8m/s^2)(sin(25))=4.142 m/s^2

Should I not be canceling the masses out?

Would a better approach be taking mgsin(25)-mgsin(40)/55 = 0.345s?

 

[andrewkirk]

The two blocks have to have the same acceleration because of the rope.

Work out the net accelerating force on the pair as the difference between the accel forces on each block - since they are pulling in opposite directions.

Then divide by the combined mass.

 

[siimplyabi]

The two blocks have to have the same acceleration because of the rope.

Work out the net accelerating force on the pair as the difference between the accel forces on each block - since they are pulling in opposite directions.

Then divide by the combined mass.

So, You're saying

a = m1gsin(theta)-m(ofblock2)gsin(theta)/m(of blk1)+m(blk2) would be the correct approach.

Could you be able to tell me when it would be appropriate to cancel out the masses? in the mgsin(theta) or mgcos(theta)= ma equation only leaving gsin(theta)= a ?

 

[siimplyabi]

So, You're saying

a = m1gsin(theta)-m(ofblock2)gsin(theta)/m(of blk1)+m(blk2) would be the correct approach.

Could you be able to tell me when it would be appropriate to cancel out the masses? in the mgsin(theta) or mgcos(theta)= ma equation only leaving gsin(theta)= a ?

Also, thank you for your response. I appreciate it.

 

[andrewkirk]

I'm afraid I don't know what you mean by 'cancel out the masses'. The masses have two different roles in such setups. First, they generate force, via gravity. Second, they are the denominator by which one divides the net force.

In this case, the contributions of the two masses are subtracted in the first role (after adjusting for the different angles) and added in the second role.

 

[siimplyabi]

I'm afraid I don't know what you mean by 'cancel out the masses'. The masses have two different roles in such setups. First, they generate force, via gravity. Second, they are the denominator by which one divides the net force.

In this case, the contributions of the two masses are subtracted in the first role (after adjusting for the different angles) and added in the second role.

Okay, thank you. I understand what you're telling me. I was just curious as to why I have seen some people refer to canceling out the masses when looking for acceleration of one object. But you've answered my question so thanks!