Finding a Mass of a Block on an Atwood's Machine

[guss]

Homework Statement

A 5.00-kg block (M) on a 30° incline is connected by a light string over a frictionless pulley to an unknown mass, m. The coefficient of kinetic friction between the block and the incline is 0.100. When the system is released from rest, mass m accelerates upward at 2.00 m/s^2. Determine (a) the string tension and (b) the value of m.

Homework Equations

F_F = μ*Normal
a = F_N/m

The Attempt at a Solution

I am going to attempt part b of this problem first because that makes the most sense to me:
To get the normal force of the block against the ramp, I did cos(30)*9.8*5 which is 42.44 N.
To get the force of the block going down the ramp neglecting friction, I did sin(30)*9.8*5, which is 24.5 N.
Then, I figured out the friction by multiplying 42.44 by .1 which is 4.244. I can then subtract 4.244 from 24.5 to get 20.256.
Now, I know that this is the force pulling up on the unknown weight that is accelerating it.
So, I know that m=F_N/a, and in this case F_N is 20.256, so I simply do 20.256/2 to get 10.128 kg, but this is not correct. I thought that might be the weight, so I divided that answer by 9.8 and it still wasn't correct.

Now for part a, I'm not sure if these equations I'm using are correct, and the answer is obviously not right because my answer to part b was not correct. But I'll give the equation I used so you guys can see what I did and help me incase I'm wrong:

m₁*(T-W₂)=m₂*(W₁-T)

I don't think that's the right equation for tension in this case, but I could be correct I suppose.

Thanks.

 

[PhanthomJay]

Homework Statement

A 5.00-kg block (M) on a 30° incline is connected by a light string over a frictionless pulley to an unknown mass, m. The coefficient of kinetic friction between the block and the incline is 0.100. When the system is released from rest, mass m accelerates upward at 2.00 m/s^2. Determine (a) the string tension and (b) the value of m.

Homework Equations

F_F = μ*Normal
a = F_N/m

The Attempt at a Solution

I am going to attempt part b of this problem first because that makes the most sense to me:
To get the normal force of the block against the ramp, I did cos(30)*9.8*5 which is 42.44 N.
To get the force of the block going down the ramp neglecting friction, I did sin(30)*9.8*5, which is 24.5 N.
Then, I figured out the friction by multiplying 42.44 by .1 which is 4.244. I can then subtract 4.244 from 24.5 to get 20.256.
Now, I know that this is the force pulling up on the unknown weight that is accelerating it.

This is not correct. You did not include the tension force in string.

So, I know that m=F_N/a, and in this case F_N is 20.256, so I simply do 20.256/2 to get 10.128 kg, but this is not correct. I thought that might be the weight, so I divided that answer by 9.8 and it still wasn't correct.

Now for part a, I'm not sure if these equations I'm using are correct, and the answer is obviously not right because my answer to part b was not correct. But I'll give the equation I used so you guys can see what I did and help me incase I'm wrong:

m₁*(T-W₂)=m₂*(W₁-T)

I don't think that's the right equation for tension in this case, but I could be correct I suppose.

Thanks.

You must draw free body diagrams of each mass, and identify the forces acting on each mass. Then the net force acting on each mass provides the acceleration of that mass, per Newton's 2nd law. Both masses accelerate at the same rate...one accelerates down the plane, the other as given accelerates up.

 

[guss]

This is not correct. You did not include the tension force in string. You must draw free body diagrams of each mass, and identify the forces acting on each mass. Then the net force acting on each mass provides the acceleration of that mass, per Newton's 2nd law. Both masses accelerate at the same rate...one accelerates down the plane, the other as given accelerates up.

I realize this. I am confused as to what steps I should take in what order to solve the problem. Should I solve for the tension or the mass of the object first? What formula(s) should I use to get this information?

 

[PhanthomJay]

You can start by drawing a free body diagram of the block on the incline. This is what you attempted to do, but you must identify all forces acting. Along the incline, there are 3 forces acting, one of which is the string tension pulling away from that mass. You can solve for that immediately, since M, a, g, u, and theta are known, using Newton 2, where the net force must act down the plane. Then draw a free body diagram of the hanging mass m. There are 2 forces acting on that mass. Identify them, and apply Newton 2 in the vertical direction to solve for m, noting that the net force must be upward, in the direction of the given acceleration.

 

[guss]

You can start by drawing a free body diagram of the block on the incline. This is what you attempted to do, but you must identify all forces acting. Along the incline, there are 3 forces acting, one of which is the string tension pulling away from that mass. You can solve for that immediately, since M, a, g, u, and theta are known, using Newton 2, where the net force must act down the plane. Then draw a free body diagram of the hanging mass m. There are 2 forces acting on that mass. Identify them, and apply Newton 2 in the vertical direction to solve for m, noting that the net force must be upward, in the direction of the given acceleration.

Ok, and to solve for the tension is where I am struggling then.

So, using M, g, u, and theta I determined that the force of the block going down the ramp is 20.256. So, is this a step in getting tension, and if so, how would I use it to get the tension?

Thanks a lot for your help.

 

[PhanthomJay]

Ok, and to solve for the tension is where I am struggling then.

So, using M, g, u, and theta I determined that the force of the block going down the ramp is 20.256. So, is this a step in getting tension, and if so, how would I use it to get the tension?

Thanks a lot for your help.

In your free body diagram of the block going down the incline, you are still forgeting to include the unknown tension force, T, acting up the plane.

You have correctly identified the weight component of 24.5 N acting down the plane.
You have correctly identified the friction component of 4.244 acting up the plane.
But you have neglected to include the tension force, T, acting up the plane.
The algebraic sum of these 3 forces is the net force acting down the plane. This net force is equal to Ma, per Newton 2, where M and a are given. Solve for T!