Finding Maximum Acceleration on a Box on a Cart with Friction

In summary, the conversation discusses a problem involving a cart on a frictionless surface with a rope attached to it. The question is what tension on the rope would give the cart a maximum acceleration. The solution involves considering the only force directly applied to the cart and using that to find the required tension in the rope. The final solution is given as T = (M+m)μsN/M, with N = mg. The correctness of the solution is confirmed with the help of an expert.
  • #1
amcavoy
665
0
This is a link to a picture of my problem:

http://img57.imageshack.us/img57/9382/cart5rh.jpg

It asks what tension on the rope would give the cart a maximum acceleration. The surface the cart is on is frictionless, while the cart itself is not (the surface the box is on).

To solve this, I figured that pulling the rope with a force of usN would give the maximum acceleration of the cart. However, this seems too easy; I think I may have not completely answered the question.

|T| = usN = usmg

Is my reasoning correct? Thanks.
 
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  • #2
You really didn't provide your reasoning, just your answer. (No, it's not correct.)

Instead, think this way: What's the only force directly applied to the cart? What's the maximum acceleration acceleration that that force can give the cart? Use that to find the required tension in the rope.
 
  • #3
What contact force is exerted on the cart?
 
  • #4
Doc Al said:
You really didn't provide your reasoning, just your answer. (No, it's not correct.)
Instead, think this way: What's the only force directly applied to the cart? What's the maximum acceleration acceleration that that force can give the cart? Use that to find the required tension in the rope.
Ok this is what I came up with:

[tex]F_M=\mu_sN\implies a_M=\frac{\mu_sN}{M}[/tex]

[tex]T=\left(M+m\right)a=\frac{\left(M+m\right)\mu_sN}{M};\quad N=mg[/tex]

Is this correct?

Thanks again.
 
  • #5
Now you've got it.
 
  • #6
Thanks Doc Al, I appreciate it.
 

Related to Finding Maximum Acceleration on a Box on a Cart with Friction

1. What is the concept of "Box on Cart (Friction)"?

The concept of "Box on Cart (Friction)" is a common physics problem that involves studying the interaction between a box placed on top of a cart and the frictional force between the two surfaces. This problem is often used to demonstrate the effects of friction on the motion of objects.

2. How is friction involved in the "Box on Cart" problem?

Friction plays a crucial role in the "Box on Cart" problem as it is the force that opposes the motion between the box and cart. The frictional force is generated due to the contact between the two surfaces and can either help or hinder the motion of the box on the cart.

3. What are the factors that affect the frictional force in the "Box on Cart" problem?

The factors that affect the frictional force in the "Box on Cart" problem include the mass of the box, the mass of the cart, the coefficient of friction between the two surfaces, and the applied force on the cart. These factors can either increase or decrease the frictional force between the box and cart.

4. How does the angle of incline affect the frictional force in the "Box on Cart" problem?

The angle of incline can significantly affect the frictional force in the "Box on Cart" problem. As the angle increases, the normal force between the box and cart decreases, resulting in a decrease in the frictional force. This means that the box will slide more easily on the cart as the angle of incline increases.

5. What are the applications of studying the "Box on Cart" problem?

The "Box on Cart" problem has several real-life applications, including understanding the physics behind pushing an object on a surface, designing better transportation systems, and improving the efficiency of machines that involve sliding or rolling objects. It also helps in understanding the role of friction in everyday life and how to reduce its effects for smoother motion.

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