# Tension on blocks on ramp

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1. Feb 28, 2016

### JustAverageSkill

1. The problem statement, all variables and given/known data
The ramp in figure is frictionless. If the blocks are released from rest, which way does the 10 kg block side and what is the magnitude of its acceleration? θ=40°

2. Relevant equations

3. The attempt at a solution
m1=10kg, m2= 5kg

For m1:
ΣFx=m1a1
T-m1gsinθ
For m2
ΣFx=m2a2
I'm unsure how to proceed from here. Other online solutions seem to assume that a1=a2, is that true? Why/why not? Please help me in how to continue. Thanks!

Last edited by a moderator: May 7, 2017
2. Feb 28, 2016

### Qwertywerty

Provided the string is in-extensible, as can be assumed here, a1 must equal a2.
The letters in bold should provide a clue to the reasoning behind this.

Hope this helps,
Qwertywerty.

3. Feb 28, 2016

### JustAverageSkill

Given that fact, how should I proceed?

4. Feb 28, 2016

### Qwertywerty

You have two unknowns - Tension and acceleration(Tension is same throughout the string).

You need two equations - Write Newton's law for both masses, and solve the two equations to obtain a.

5. Feb 29, 2016

### JustAverageSkill

I tried to write out the equations as follows:
m1ax=T - m1gsinθ
m2ax=T - m2g

Is this correct? Upon solving this I got an answer that is different than the one in the book

6. Feb 29, 2016

### Qwertywerty

Edit : Keep in mind, the directions of both accelerations. We have already arrived at a conclusion regarding the accelerations. What is that?

7. Feb 29, 2016

### JustAverageSkill

I'm not sure what you mean by this. For m1, I have defined positive x to be up and parallel to the ramp. For m2, positive x is directly up.

8. Feb 29, 2016

### Qwertywerty

If one block goes up, the other one goes...Where?

9. Feb 29, 2016

### JustAverageSkill

If one block goes up, the other goes down, I suppose. But how do I quantitate that idea?

10. Feb 29, 2016

### JustAverageSkill

11. Feb 29, 2016

### Qwertywerty

For one block, acceleration would be in the direction of tension; in the other, not. You have to thus be careful while applying Newton's law.

12. Feb 29, 2016

### JustAverageSkill

Could you please write out the correct force equations that I should use to solve for tension and acceleration?

13. Feb 29, 2016

### Qwertywerty

I'd rather you tried it for yourself.

Assume that the 10kg mass goes up and the 5kg one down. Try the equations again.

P.S - I have noted that you had made an attempt in post #5, but I'd like for you, to give it another go.

14. Feb 29, 2016

### JustAverageSkill

Here's another attempt, which yielded the right answer.
For m1:
m1ax=T-m1gsinθ
For m2:
m2(-ax)=T-m2g
m1ax+m1gsinθ=m2(-ax)+m2g
ax(m1+m2)=m2g-m1gsinθ
ax=(m2g-m1gsinθ)/(m1+m2)=(5*9.8-10*9.8*.64)/15=-0.92

15. Feb 29, 2016

### Qwertywerty

That seems to be correct. Congrats!