At what angle should I hold a rope to maximize the horizontal acceleration of a crate?

Question

38 kg crate is being dragged across a horizontal floor by means of a rope held at an angle of 36 degrees with the horizontal. The coefficient of friction between the crate and the floor is 0.22 and the tension in the rope is 800 N. What angle should one hold the rope at to maximize the horizontal acceleration of the crate?

Thanks for any help! =)

Answer 1

We assume that the coefficient K=0.22 is the coefficient of kinetic friction (static friction is only relevant when the thing does not move yet, which is not what we're faced with here).

k=0.22 = F/N is the ratio of the (horizontal) friction force F to the total normal force N (the sum of all vertical forces).

Let u be the angle the rope of tension T = 800 N makes with the horizontal.  The normal force N is the weight (Mg) of the crate minus the upward component of the force exerted by the rope:

   N = Mg - T sin(u)

The horizontal force is the horizontal component of the pull from the rope minus the friction force F = kN (with the above expression for N).

T cos(u) - k N  =  T cos(u) - k (Mg - T sin(u) )

We want this expression to be maximum (so that the acceleration of the crate will be maximum). So, its derivative with respect to u must vanish, which is to say:

0 = - T sin(u) + k T cos(u)

In other words: k = tan(u)

So, u = arctan(0.22) = 12.4 degrees

Normally, one final check SHOULD normally be made in this type of problem to ensure that the forces are sufficient to overcome friction to begin with.  This is not necessary here because we are told implicitely that the crate is ALREADY moving. The above solution would thus be valid even if the tension of the rope was only 1 N, although that case would clearly mean that the crate would decelerate and soon come to a halt (we're indeed asked to maximize acceleration, which may mean to make the absolute value of a negative acceleration as small as possible).

Nevertheless, it's nice to check that, with the numbers given, the acceleration is clearly positive.  Its maximal value being

a = [T cos(u) - k (Mg - T sin(u) )] / M

where cos(u)=1/sqrt(1+k^2) and sin(u)=k/sqrt(1+k^2)

This boils down to:

a = sqrt(1+k^2) T/M - k g

= sqrt(1.0484) (800/38) - 0.22 (9.80665) = 19.3986 m/s^2

By comparison, with an horizontal rope (u=0) we would have:

a = T/M - k g = 800/38 - 0.22 (9.80665) = 18.8952 m/s^2

which is, indeed, slightly less.

Answer 2

horizontal. the maximum friction is 22% of the crate's gravitational force.