What Is the Tension Required for the Box to Slip on a Snow-Covered Hill?

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The discussion revolves around calculating the tension required for a box to slip on a sled on a snow-covered hill at a 22-degree incline. The sled weighs 18 kg, and the box weighs 9 kg, leading to a total weight that contributes to the forces acting on the system. Initial calculations suggest a tension of approximately 99 N based on the weight component down the slope, but this is deemed incorrect due to the absence of friction information. The need for the coefficient of friction between the box and sled is highlighted as essential for accurate calculations. Without this friction value, the box will slide if there is no friction present.
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Homework Statement


A rope attached to a 18.0kg wood sled pulls the sled up a 22.0 degree snow-covered hill. A 9.00kg wood box rides on top of the sled.

Mass of Sled: 18kg
Mass of Box: 9kg
Angle of Incline: 22 Deg

If the tension in the rope steadily increases, at what value of the tension does the box slip?

Homework Equations


F=ma

The Attempt at a Solution


So I tried getting the overall tension that keeps the sled and box in equilibrium. So the tension is equal to the weight component down the slope which would be (27)(9.8)(sin 22). That gives that the tension is ~99N. So anything over that would make it slip but nope... apparently it's wrong.
 
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Consider the friction between box and sled. (Looks like information is missing from your problem statement.)
 
The friction is not given in the problem.
 
octopus41092 said:
The friction is not given in the problem.
If there's no friction between box and sled, the box will slide. Does your book give a coefficient of friction for wood on wood?
 
you need the coefficient of friction between box and sledge and then work with fbd of box and sled
 
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