Max Tension Force for Box Not to Slip: 24.5N

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SUMMARY

The maximum tension force that prevents a 5.0 kg wood box from slipping on a 10 kg sled is calculated to be 24.5 N. This value is derived using the static friction formula, where the coefficient of static friction (μ) for wood on wood is 0.5. The calculation involves determining the normal force acting on the box, which is equal to its weight (5 kg * 9.81 m/s²). Additionally, the sled's mass is relevant as it also contributes to the overall acceleration when the tension force is applied.

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Homework Statement



A horizontal rope pulls a 10kg wood sled across frictionless snow. A 5.0kg wood box rides on the sled. The coefficient of static friction for wood on wood is mu = 0.5.

What is the largest tension force for which the box doesn't slip?

Homework Equations



f(static) = mu * f(normal)


The Attempt at a Solution



I don't really see how the mass of the sled is relevant. But here's what I did:

f(static) = 0.5 * 5(9.81) = 24.5N
 
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You've calculated the maximum horizontal force the sled can exert on the 5 Kg box. That is not the whole answer to the question though.

In order to exert that force and accelerate the box, the sled itself must also be accelerating, and therefore you must calculate the acceleration that the 5 Kg box would undergo with a force of 24.5 N acting on it. The max tension in the rope would be the force required to accelerate the box AND the sled.
 
ah ok. Thanks for the reply.
 

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