What is the acceleration of the 10 kg block on the frictionless ramp?

In summary: For m1:m1ax=T-m1gsinθFor m2:m2(-ax)=T-m2gm1ax+m1gsinθ=m2(-ax)+m2gax(m1+m2)=m2g-m1gsinθax=(m2g-m1gsinθ)/(m1+m2)=(5*9.8-10*9.8*.64)/15=-0.92That seems to be correct. Congrats!
  • #1
JustAverageSkill
8
0

Homework Statement


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°
https://s.yimg.com/hd/answers/i/b94e08c806334d908a6afee7b8761bbd_A.png?a=answers&mr=0&x=1456730351&s=946fcf963764a48cb25be59052f47790

Homework Equations

The Attempt at a Solution


m1=10kg, m2= 5kg[/B]
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!
 
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  • #2
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
Qwertywerty said:
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.
Given that fact, how should I proceed?
 
  • #4
JustAverageSkill said:
Given that fact, how should I proceed?
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
Qwertywerty said:
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.
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
Edit : Keep in mind, the directions of both accelerations. We have already arrived at a conclusion regarding the accelerations. What is that?
 
  • #7
Qwertywerty said:
Keep in mind, the directions of both accelerations. Assume one common direction for both(Why?).
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
JustAverageSkill said:
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.
If one block goes up, the other one goes...Where?
 
  • #9
If one block goes up, the other goes down, I suppose. But how do I quantitate that idea?
 
  • #10
Qwertywerty said:
If one block goes up, the other one goes...Where?
See above please
 
  • #11
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
Qwertywerty said:
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.
Could you please write out the correct force equations that I should use to solve for tension and acceleration?
 
  • #13
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
Qwertywerty said:
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.
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
That seems to be correct. Congrats!
 

Related to What is the acceleration of the 10 kg block on the frictionless ramp?

What is tension on blocks on ramp?

Tension on blocks on ramp refers to the force that is exerted by a rope or cable on a block that is resting on an inclined plane or ramp. This force is a result of the weight of the block and the angle of the ramp, and it is responsible for keeping the block in place and preventing it from sliding down the ramp.

How is tension calculated on blocks on ramp?

Tension can be calculated using the formula T = Wsinθ, where T is the tension, W is the weight of the block, and θ is the angle of the ramp. This formula takes into account the vertical and horizontal components of the weight of the block and the angle of the ramp.

How does the angle of the ramp affect the tension on blocks?

The angle of the ramp has a direct impact on the tension on blocks. As the angle increases, the tension also increases, as there is a greater vertical component of the weight of the block. On the other hand, as the angle decreases, the tension decreases, as there is a smaller vertical component of the weight of the block.

What happens to the tension if the ramp is frictionless?

If the ramp is frictionless, the tension on blocks will be equal to the weight of the block. This is because there is no force acting against the block's weight, and therefore, the tension will only need to counteract the weight to keep the block in place.

How is tension affected by the weight of the block?

The weight of the block directly affects the tension on blocks on ramp. As the weight of the block increases, the tension also increases, as there is a greater force pulling the block down the ramp. Conversely, as the weight of the block decreases, the tension also decreases.

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