What angle should movers set their truck ramp to minimize work?

[tricky_tick]

Movers want to set the ramp of their truck so that the work they do against the combination of gravity and friction is a minimum for crates moving up the ramp with constant velocity. µ is the coefficient of kinetic friction and θ is the angle between the ramp and the ground. For the work to be a minimum, they must choose:

a. tan θ = µ

b. tan θ = -µ

c. tan θ = -1/µ

d. tan θ = 1/µ

e. tan θ = 1 - µ

The obvious solution of setting the derivative of work equal to zero finds the maximum work. The problem asks for the minimum?

 

[HallsofIvy]

The word done is "Force*distance".

The force necessary to move an object (I'm assuming they are sliding it) is the component of weight along the ramp: mg sin(θ) (draw a picture and look at the right triangles!) plus the friction force: μ times the component of weight perpendicular to the ramp, mg cos(θ).
Force= mg sin(θ)+ mg μ cos(θ).

The distance is the length of the hypotenuse of the right triangle formed by the ramp: h/cos(θ) where h is the height of the ramp (which I am assuming is fixed: the height of the truck bed).

The work done is hmg(sin(θ)+ μcos(θ)/cos(θ).

What value of θ minimizes that?

 

[wais]

The correct answer is a. tan θ = µ. This is because when the derivative of work is set to zero, it finds the maximum work. However, in this scenario, we want to minimize the work, which means we need to find the angle that will result in the least amount of work. By setting tan θ = µ, we are finding the angle that balances the force of gravity and friction, resulting in the minimum amount of work required to move the crates up the ramp with constant velocity. Therefore, movers should set their truck ramp at an angle where the tangent of the angle is equal to the coefficient of kinetic friction. This will minimize the work they have to do against gravity and friction, making their job more efficient.