Physics: Inclined Plane With Wheels

AI Thread Summary
The discussion revolves around a physics problem involving a 4 kg block on a 37-degree inclined plane with wheels, where the goal is to determine the forces acting on the system under various conditions. In part A, the required force to accelerate the system at 1.5 m/s² is calculated to be 30 N, and there are questions about the equations used to find the tension in the rope. Part B raises confusion regarding how a horizontal force can reduce the normal force on the block to zero, with participants seeking clarification on the implications of a frictionless surface. The presence of static and sliding friction coefficients adds complexity to the problem, leading to further inquiries about the setup and forces involved. Overall, the thread highlights the challenges of understanding the dynamics of the system and the role of friction in inclined plane problems.
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Homework Statement



A 4 kg block rests on 37 degrees inclined plane with wheels, held in place by a light rope which is parallel to the plane. The coefficient of static friction is 0.3 and the coefficient of sliding friction is 0.2. The mass of the inclined plane (with wheels) is 16 kg.

A.) The block is attached with the rope to the incline and the surfaces are treated with a magical substance that eliminates friction. The wheels are unlocked, and a horizontal force is applied to the plane so that it rolls and accelerates to the right at 1.5 m/s/s.

I. Find the magnitude of the force required to the make the system accelerate.
II. Find the tension in the rope.

B.) With the conditions in part A, find the minimum horizontal force applied to the apparatus that reduces the normal force of the plane on the 4 kg block to zero.

Homework Equations



F=ma
ƒ(friction)=\muN


The Attempt at a Solution



A.) For I.) F_x = F=ma so F=(16 + 4)(1.5) = 30 N
Is the question asking how hard to push the whole inclined plane with the box in order for it to accelerate at 1.5 m/s/s?

For II.) F_y=N-mgcos\theta=0 so N=mgcos\theta
and F_x=T-mgsin\theta=ma so T=mgsin\theta+ma
Are these the correct equations to use?


B.) ƒ=\muN and ƒ=0 but I don't really understand what the question is asking. The normal force on the box would be N=mgcos\theta, but how could a horizontal force make a vertical force equal to zero.

Thanks
 
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Can you draw the figure?
In the statement of the problem you have given the μs and μk.
But in A you mention that the surface is frictionless. What it means? I can't imagine the position of the rope.
 
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