A heavy sled is being pulled by two people as shown in the figure. The coefficie

Question

A heavy sled is being pulled by two people as shown in the figure. The coefficient of static friction between the sled and the ground is s = 0.603, and the kinetic friction coefficient is k = 0.419. The combined mass of the sled and its load is m = 321 kg. The ropes are separated by an angle = 28°, and they make an angle = 31.1° with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving?

If this rope tension is maintained after the sled starts moving, what is the sled's acceleration?

Explanation / Answer

As both ropes pull equally hard, the tension in both the ropes remain same

To make the sled start moving, the horizontal tension component of the ropes should balance out the static friction.

First let's find the horizontal planar components of the two tensions.

Let the tensions in the rope be T

As the rope makes 31.1 degrees with the horizontal, their planar components will be T cos(31.1) and T cos(31.1)

Now, these two components are separated by angle 28 degrees. The components of the above planar components opposing the static friction will be T cos(31.1) cos(14) , T cos(31.1) cos(14)

(As, there is no diagram, I am assuming the symmetry of the two ropes along the line of frictional force. Hence, 28/2 will be the angle the tensions make to the line of frictional force)

Now, equating forces

2 * T * cos(14)* cos(31.1) = mus * m * g

2 * T * 0.97 * 0.856 = 0.603 * 321 * 9.81

=> T = 1143.45 N

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