Finding Maximum Acceleration on a Box on a Cart with Friction

AI Thread Summary
The discussion centers on determining the tension in a rope that would provide maximum acceleration to a cart on a frictionless surface, while the box on the cart experiences friction. The initial reasoning proposed that pulling with a force equal to the static friction would yield maximum acceleration, but this was deemed incorrect. Participants emphasized focusing on the forces directly applied to the cart and how they relate to the maximum acceleration achievable. The final formulation provided by the user correctly incorporates the static friction coefficient and mass to calculate the required tension in the rope. Overall, the conversation highlights the importance of understanding the dynamics of forces in a friction-influenced system.
amcavoy
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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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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.
 
What contact force is exerted on the cart?
 
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:

F_M=\mu_sN\implies a_M=\frac{\mu_sN}{M}

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

Is this correct?

Thanks again.
 
Now you've got it.
 
Thanks Doc Al, I appreciate it.
 
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