Physics: Inclined Plane With Wheels

[passphysics]

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)=[tex]\mu[/tex]N

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[tex]\theta[/tex]=0 so N=mgcos[tex]\theta[/tex]
and F_x=T-mgsin[tex]\theta[/tex]=ma so T=mgsin[tex]\theta[/tex]+ma
Are these the correct equations to use?


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

Thanks

 

[rl.bhat]

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.

 

[tomcue]

for your question. I can provide a response to the content and help clarify any confusion or questions you may have.

For part A, your attempt at a solution is correct. The question is asking for the magnitude of the force required to make the entire system (inclined plane and block) accelerate at 1.5 m/s/s. This force would be applied horizontally to the incline.

For II, your equations are also correct. The tension in the rope would be equal to the sum of the weight of the block (mg) and the force required to accelerate it (ma), which is acting in the same direction as the force of gravity. This makes sense since the rope is parallel to the incline and is not contributing to the normal force.

For part B, the question is asking for the minimum horizontal force that can be applied to the apparatus (inclined plane and block) in order to reduce the normal force on the block to zero. This means that the block would be on the verge of sliding down the incline without any additional force. In this case, the frictional force would also be zero, so the only forces acting on the block would be the weight (mg) and the horizontal force. You can use the equation F=ma to solve for the minimum force required.