Which Ramp is Easier to Raise a 2kg Object? Find Out!

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The discussion centers on determining which ramp is easier to raise a 2kg object and why. It concludes that the second ramp requires less force (40N) compared to the first ramp (100N) due to the longer distance the object must be moved. However, questions arise regarding the acceleration of the mass when using a smaller force against gravity's pull of 200N. The conversation emphasizes the importance of understanding the components of forces acting on the object, particularly the parallel and perpendicular components related to the incline. Ultimately, the ramp's normal force plays a crucial role in balancing the gravitational force, allowing the object to be moved without accelerating downward.
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Homework Statement


A ramp is shown in the picture.Which ramp is easier to rise an object of mass 2kg up?and why?


Homework Equations


W=FS



The Attempt at a Solution


The second ramp.The work done in rising the mass against gravity is same in the two ramps.That is
W=200N*1 = 200J
But rising in the second ramp,the mass has to move more distance.
200J=5*F So less force is needed F=40N
The first ramp
200J=2*F F=100N
In the second ramp one has to use less force to move it up so it is easier.
My question is:If he uses 40N against gravity(200N),Why doesn't the mass accelerate down?
 

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What does 'easier' mean exactly?

adjacent said:
The second ramp.The work done in rising the mass against gravity is same in the two ramps.That is
W=200N*1 = 200J
If the mass is at rest before and after its displacement, the work done by the vertical component of your applied force, call it Wv, is the same in each case (and is equal in magnitude to the work done by gravity).

Can you show the calculation that gave you a weight of 200 N?

adjacent said:
But rising in the second ramp,the mass has to move more distance.
200J=5*F So less force is needed F=40N
The first ramp
200J=2*F F=100N
In the second ramp one has to use less force to move it up so it is easier.
My question is:If he uses 40N against gravity(200N),Why doesn't the mass accelerate down?
Here your applied force acts on the mass in a direction parallel to the incline of the ramp. The vertical component of this force does work that's less in magnitude than Wv in moving the mass up the ramp. If no other force besides your applied force and gravity act on the mass, then you have described a situation that violates the law of conservation of energy.
 
Last edited:
milesyoung said:
Can you show the calculation that gave you a weight of 200 N?
Simple F=ma
F=2*10(-Acceleration due to gravity):smile:

milesyoung said:
You have described a situation that violates the law of conservation of energy.
Why?
 
milesyoung said:
Can you show the calculation that gave you a weight of 200 N?
Simple F=ma
F=2*10(-Acceleration due to gravity):smile:

milesyoung said:
You have described a situation that violates the law of conservation of energy.
Why?Can you explain your answer.
Sorry.I am not able to delete the first one!
 
adjacent said:
Simple F=ma
F=2*10(-Acceleration due to gravity):smile:
And what is 2*10 equal to?

adjacent said:
Why?Can you explain your answer.
The work done on the mass by the force that acts in a direction parallel to the incline of the ramp is the sum of the work done by its horizontal and vertical components. You're saying this sum is equal to 200 J. If you know the work done by the vertical component has to be 200 J for the mass to move to its final position at the top of the ramp, then the work done by the horizontal component has to be 0 J. Since your applied force has a nonzero horizontal component, the mass can't move to the top of the ramp as described in your post without violating the law of conservation of energy.
 
milesyoung said:
And what is 2*10 equal to?
Actually it was a spelling mistake,It should be 20kg:rolleyes:

milesyoung said:
The work done on the mass by the force that acts in a direction parallel to the incline of the ramp is the sum of the work done by its horizontal and vertical components. You're saying this sum is equal to 200 J. If you know the work done by the vertical component has to be 200 J for the mass to move to its final position at the top of the ramp, then the work done by the horizontal component has to be 0 J. Since your applied force has a nonzero horizontal component, the mass can't move to the top of the ramp as described in your post without violating the law of conservation of energy.
I was talking about the work done by gravity(That is vertical component)
 
adjacent said:
I was talking about the work done by gravity(That is vertical component)
I don't follow. Could you be more specific?
 
milesyoung said:
I don't follow. Could you be more specific?
This question was based on a question on my textbook which is as follows:
(In the picture)
A trolley weights 500N.What force F is needed to pull it up From A to C,assuming that there is no friction?
Answer(Given in the textbook itself)
To raise the trolley vertically from B to C,a force of 500N is needed.
So:Work done=Force x distance moved= 500N x 2m =1000J
Treating the ramp as a ligting machine,this is its useful work output.

If the trolley is pulled up the ramp from A to C.
Work done= Force X distance moved = F X 4m=1000J
This is work output

As there is no friction ; Useful work output=work input
So,leaving out the units for simplicity:F x 4=1000
This gives F= 250N,so the force needed to pull the trolley is 250N.
That answer did not refer to horizontal works done.It merely relied on vertical work.This is similar to my question><

Can you give your answer in the way you were giving it before completely?
 

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adjacent said:
My question is:If he uses 40N against gravity(200N),Why doesn't the mass accelerate down?
What's the component of gravity acting down the incline?
 
  • #10
doc al said:
what's the component of gravity acting down the incline?
200Newton
 
  • #11
adjacent said:
200n
No. 200 N is the total gravitational force, which acts downward. You need the component acting down the incline (parallel to its surface).
 
  • #12
Doc Al said:
No. 200 N is the total gravitational force, which acts downward. You need the component acting down the incline (parallel to its surface).
What is a component?
 
  • #13
adjacent said:
What is a component?
Forces, like weight, are vectors. Have you studied vectors?

Read this about Inclined Planes.
 
  • #14
Doc Al said:
Forces, like weight, are vectors. Have you studied vectors?

Read this about Inclined Planes.
I know about vectors but not those about inclined planes.However,the question does not give any information about any components
 
  • #15
adjacent said:
I know about vectors but not those about inclined planes.However,the question does not give any information about any components
Not directly, but you can figure it out. In fact, that's what you did, in effect.

Without the benefit of a ramp, you must exert the full weight of 200 N upward to lift the mass 1 meter.

But with a ramp, the force you have to exert needs only to balance the component of the weight acting down the ramp. (The normal force of the ramp surface does the rest.) For the first ramp, that component is 100 N. For the second, it's 40 N. Of course, the lower the force you have to exert, the longer a distance you need to push the mass to get it to the same height. The work done is the same.
 
  • #16
Doc Al said:
Not directly, but you can figure it out. In fact, that's what you did, in effect.

Without the benefit of a ramp, you must exert the full weight of 200 N upward to lift the mass 1 meter.

But with a ramp, the force you have to exert needs only to balance the component of the weight acting down the ramp. (The normal force of the ramp surface does the rest.) For the first ramp, that component is 100 N. For the second, it's 40 N. Of course, the lower the force you have to exert, the longer a distance you need to push the mass to get it to the same height. The work done is the same.
Then is the answer to my question is the normal force of the ramp supplies the rest of 160N force for the second ramp?
 
  • #17
adjacent said:
Then is the answer to my question is the normal force of the ramp supplies the rest of 160N force for the second ramp?
The normal force "cancels" the component of gravity perpendicular to the surface, but that component is not 160 N (it's about 196 N). All that's left is the component parallel to the surface, which is what you have to exert to push it up the ramp. That parallel component is 40 N.
 
  • #18
Doc Al said:
The normal force "cancels" the component of gravity perpendicular to the surface, but that component is not 160 N (it's about 196 N). All that's left is the component parallel to the surface, which is what you have to exert to push it up the ramp. That parallel component is 40 N.

Can I assume the ramp supplies the rest of the force the I does not have to(That is I have to exert a force of 40N,If the ramp does not exert the rest of the (160)Newton of force against weight,it would accelerate down?
 
  • #19
adjacent said:
Can I assume the ramp supplies the rest of the force the I does not have to(That is I have to exert a force of 40N,If the ramp does not exert the rest of the (160)Newton of force against weight,it would accelerate down?
Not exactly sure what you're asking. Since we presume that the mass is being raised at some constant speed, the net force on it must be zero. You are exerting a force of 40 N at an angle; the vertical component of your force is only about 8 N. The ramp surface exerts the other 192 N of vertical force needed to support the weight.
 
  • #20
Doc Al said:
the vertical component of your force is only about 8 N.
Does that mean we are exerting a force of about 8N against gravity?
 
  • #21
adjacent said:
Does that mean we are exerting a force of about 8N against gravity?
I would not put it that way. I'd say, as I already have, that you are exerting a force equal to the component of gravity acting down the ramp, which equals 40 N. (The ramp takes care of the component of gravity acting perpendicular to the ramp.)

(There are many ways to look at this.)
 
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