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## Homework Statement

A block of mass m is pulled across a level surface by a rope that makes an angle θ with the horizontal. The coefficient of friction is μ.

(a) Determine the amount of force F required to slide the block at a constant velocity.

(b) Determine the optimum angle at which to pull on the block (so that the required force is minimized).

(c) If the force of the rope is 15.0 N acting on a block of mass 2.00 kg where μ = 0.35, what is the maximum acceleration possible?

## Homework Equations

Net force in x and y directions; Fk = [tex]\mu[/tex]Fn; vector components

## The Attempt at a Solution

a) The solution to part a) is found by combining the information gained when writing the net force equations in the x and y directions. The final, simplified solution for the force required to slide the block at constant velocity is F = ([tex]\mu[/tex]mg)/(cos[tex]\theta[/tex] + [tex]\mu[/tex]sin[tex]\theta[/tex])

b) Not sure...To find the optimum angle (presumed to correspond to min force) I think I would want to find the derivative of my solution (wrt theta) to the force equation found in part a) and set the derivative equal to zero and then solve for theta. I have been unsuccessful in doing this thus far because the derivative I found for the force equation is complex and not easy to solve for [tex]\theta[/tex] when set equal to zero. Help!

c) I would assume you could find the maximum acceleration by finding the maximum force which would be accomplished by finding the derivative as in part b), but since I've had little success there, I'm stuck here. One might assume that the max acceleration would occur at an angle of zero, but this is not necessarily true I believe because any verticle force component would reduce the frictional force, thus increasing the net acceleration of the box. Regardless, I need help here!

Finally, the answers to parts a, b, and c are listed below. Alas, while I have the answers, it is the solutions I seek. Please help :)

a. F = ([tex]\mu[/tex]mg)/(cos[tex]\theta[/tex] + [tex]\mu[/tex]sin[tex]\theta[/tex])

b. [tex]\theta[/tex] = arctan([tex]\mu[/tex])

c. 4.5 m/s2, 0°

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