Law of Motion: Horizontal Force

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Homework Help Overview

The discussion revolves around Newton's laws of motion, specifically focusing on a scenario involving two boxes with different masses placed on a horizontal surface. Participants are exploring the effects of applied forces and tension on the acceleration of the boxes and the net forces acting on them.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to calculate the acceleration of the 2 kg box based on the forces applied and are questioning how tension affects the overall system. There are discussions about whether the tension should be considered in the net acceleration of the boxes and how the forces interact.

Discussion Status

Several participants are actively engaging with the problem, raising questions about the assumptions made regarding the forces and accelerations. Some guidance has been offered regarding the relationship between the forces acting on the boxes and the implications of Newton's Third Law, but no consensus has been reached on the calculations or interpretations.

Contextual Notes

Participants express confusion regarding the application of forces and the resulting accelerations, indicating a need for clarification on the concepts of net force and tension in the context of the problem. There is also mention of the lack of friction, which may influence the dynamics being discussed.

aokidopi
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Physics: Newton Law of Motion Question?
I am a bit confused about this elementary question. If anyone can guide me in the right direction, I would appreciate it.

Picture: http://tinypic.com/r/14w84g1/7"

2 boxes are placed next to 1 another on a horizontal surface. The box on the left has a
mass of 5 kg while the box on the right has a mass of 2 kg. Assume that there is no friction
is negligible. By pulling on a horizontal string attached to the 2 kg box, Stephane is exerting a force T to the right, while you are exerting a force F to the right by pushing on the 5 kg box.

Using F = 12 N and T = 2.0 N,

(a) Determine the acceleration of the 2 kg box.

[ There is an acceleration exerted on the 2 kg box by the 5 kg. Is this 12N/ 5 kg or 2.4 m/s^2. The tension applies solely to the 2kg box and is 2 N/2 kg or 1 m/s^2. The net acceleration should be 3.4 m/s^2. If that is the case, then the 2 kg box would be pulled by itself. Thus is the total net acceleration 2.4 m/s^2 because that acceleration outweighs the tension on the 2 kg box?]


(b) Which box experiences the largest net force in this situation?

[Should be the same based on part a?]

(c) Determine the magnitude of the force exerted by the 2 kg box on the 5 kg box
in this situation.

If F = 10.0 N and T = 10.0 N, (d) Determine the acceleration of each of the boxes now.

Thank you in advance.
 
Last edited by a moderator:
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aokidopi said:
Physics: Newton Law of Motion Question?
I am a bit confused about this elementary question. If anyone can guide me in the right direction, I would appreciate it.

Picture: http://tinypic.com/r/14w84g1/7"

2 boxes are placed next to 1 another on a horizontal surface. The box on the left has a
mass of 5 kg while the box on the right has a mass of 2 kg. Assume that there is no friction
is negligible. By pulling on a horizontal string attached to the 2 kg box, Stephane is exerting a force T to the right, while you are exerting a force F to the right by pushing on the 5 kg box.

Using F = 12 N and T = 2.0 N,

(a) Determine the acceleration of the 2 kg box.

[ There is an acceleration exerted on the 2 kg box by the 5 kg. Is this 12N/ 5 kg or 2.4 m/s^2. The tension applies solely to the 2kg box and is 2 N/2 kg or 1 m/s^2. The net acceleration should be 3.4 m/s^2. If that is the case, then the 2 kg box would be pulled by itself. Thus is the total net acceleration 2.4 m/s^2 because that acceleration outweighs the tension on the 2 kg box?]


(b) Which box experiences the largest net force in this situation?

[Should be the same based on part a?]

(c) Determine the magnitude of the force exerted by the 2 kg box on the 5 kg box
in this situation.

If F = 10.0 N and T = 10.0 N, (d) Determine the acceleration of each of the boxes now.

Thank you in advance.

Firstly I am worried by this statement of yours:
"There is an acceleration exerted on the 2 kg box by the 5 kg."

A force may be exerted , but not an acceleration!

1). If the sole force acting was Stephane's, the 2kg box would accelerate away from the 5 kg box - you can easily calculate that rate of acceleration.
2). If the sole force acting was the Force you applied, then you would be accelerating the 7 kg combination. You could calculate that acceleration.
But both forces are applied:
If Stephane's force had really accelerated the 2 kg box away from the 5kg box, then Your force would only have to accelerate the 5kg box - giving a higher acceleration than you will have calculated in 2)., above.
If in that scenario, the 5kg box would be accelerating at a higher rate than the 2kg box [an impossibility] then effectively the 7kg combination is being accelerated by the combined force of you and Stephane.
 
Last edited by a moderator:
I am not sure of the last step you mentioned.

a). The acceleration of 2.0 kg box. I tried to use a few household objects to understand this. If the T= 2 N on the 2 kg box and the F=12 N, in the picture, then the tension applied on box 2kg would be canceled, would it not? The whole system is pushed by a 12 N force. The acceleration of the system is 12 N/ (2kg +5kg) = 1.71 m/s^2. The tension on box 2kg is 2 N and causes the box to accelerate at 1 m/s^2. Should these forces be added for the net acceleration on box 2kg, 2.71 m/s^2, or is it simply the acceleration of the whole system, 1.71 m/s^2.

b.) Which experiences the largest net force. The the whole system experiences the same acceleration, don't both boxes have the same net force. However, this doesn't seem correct to me. Is the Force T not taken into account here at all. That 1m/s^2 acceleration or 2 N force of and on the box 2 kg does not matter? I am rather confused.

c.) Is this simply 10 N, F-T=10 N. Because the T is a smaller magnitude, while F is a larger magnitude, box 2 kg stands in the way of the 5 kg box.

If anyone can explain this a bit more clearly, I would be thankful and less frustrated with myself.
 
bump 10char
 
aokidopi said:
I am not sure of the last step you mentioned.

a). The acceleration of 2.0 kg box. I tried to use a few household objects to understand this. If the T= 2 N on the 2 kg box and the F=12 N, in the picture, then the tension applied on box 2kg would be canceled, would it not? The whole system is pushed by a 12 N force. The acceleration of the system is 12 N/ (2kg +5kg) = 1.71 m/s^2. The tension on box 2kg is 2 N and causes the box to accelerate at 1 m/s^2. Should these forces be added for the net acceleration on box 2kg, 2.71 m/s^2, or is it simply the acceleration of the whole system, 1.71 m/s^2.

b.) Which experiences the largest net force. The the whole system experiences the same acceleration, don't both boxes have the same net force. However, this doesn't seem correct to me. Is the Force T not taken into account here at all. That 1m/s^2 acceleration or 2 N force of and on the box 2 kg does not matter? I am rather confused.

c.) Is this simply 10 N, F-T=10 N. Because the T is a smaller magnitude, while F is a larger magnitude, box 2 kg stands in the way of the 5 kg box.

If anyone can explain this a bit more clearly, I would be thankful and less frustrated with myself.

The Tension would not be canceled out by a force in the same direction! Augmented perhaps.

You push the 5kg box with a force of 12 N to the right.
The 2 kg box pushes the 5 kg box with a force I will call F2 to the left.
Net force is 12 - F2 to the right.

Steph pulls the 2kg box to the right with a force of 2N
The 5 kg box pushes the 2 kg box with a force F2 to the right. [Newtons 3rd law couple]
Net force on 2kg box is 2 + F2

Those forces result in an acceleration of the blocks, [12-F2 accelerates the 5kg box, 2+F2 accelerates the 2kg box], and the acceleration is the same for each box.

Simultaneous equations, solve for a and the contact force F2
 
Thank you very much for your help, PeterO. I think the trouble I had was grasping on the idea that the contact forces from A to B is the same as B to A. Thanks again.
 
aokidopi said:
Thank you very much for your help, PeterO. I think the trouble I had was grasping on the idea that the contact forces from A to B is the same as B to A. Thanks again.

That last bit - the A to b, B to A, equal forces - is Newton's Third law. From experience, approx 3% of Senior High students have a confident grasp of Newton's Third law.
 

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