What is the optimal angle for minimizing work on an inclined plane?

[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?

 

[marlon]

THIS IS UNTRUE. First derivative = 0 does NOT automatically mean that you are calculating the maximal value... If this were the case then the first derivative is positive on BOTH sides of the "zero-point".

Please do not double post...

https://www.physicsforums.com/showthread.php?p=362087#post362087

marlon

 

[wais]

The optimal angle for minimizing work on an inclined plane is when the tangent of the angle (θ) is equal to the coefficient of kinetic friction (µ). This can be represented as option a, tan θ = µ. This means that the angle of the ramp should be such that the tangent of the angle is equal to the coefficient of kinetic friction between the ramp and the crates being moved.

To understand why this is the optimal angle, we need to consider the forces acting on the crates as they are being moved up the ramp. These forces include the force of gravity pulling the crates downwards and the force of friction acting against the movement of the crates. As the angle of the ramp increases, the force of gravity pulling the crates downwards also increases. This means that more work needs to be done to overcome the force of gravity and move the crates up the ramp.

At the same time, the force of friction also increases as the angle of the ramp increases. This is because the component of the weight of the crates acting parallel to the ramp also increases with the angle. This means that more work needs to be done to overcome the force of friction as well.

By setting the tangent of the angle equal to the coefficient of kinetic friction, we are essentially finding the angle at which these two forces are balanced. This results in the minimum amount of work needed to move the crates up the ramp with constant velocity. Any other angle would either require more work or result in the crates moving at a non-constant velocity.

Therefore, the optimal angle for minimizing work on an inclined plane is when the tangent of the angle is equal to the coefficient of kinetic friction. This can be represented by option a, tan θ = µ. Movers should set their ramp at this angle to minimize the work they need to do against the combination of gravity and friction when moving crates up the ramp with constant velocity.