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A heavy sled is being pulled by two people. The coefficient of static friction b

ID: 1960421 • Letter: A

Question

A heavy sled is being pulled by two people. The coefficient of static friction between the sled and the ground is s = 0.595 and the kinetic friction coefficient is k = 0.403. The combined mas of the sled and its load is m =276 kg. The ropes are separated by an angle = 27 degrees, and they make and angle of = 30.8 degrees with the horizontal. assuming both ropes pull equally hard, a) what is the minimum rope tension required to get the sled moving? and b) if the tension is maintained after the sled starts moving, what is the sled's acceleration?

Explanation / Answer

To get the sled moving, the horizontal components of the tension forces must be equal to the static force.

Horizontal components of the tension forces = T * cos 30.8
To determine the component of these forces that are parallel to the direction the sled is moving, multiply be cos 24.

T * cos 30.8 * cos 27
Since there are two ropes, the total force is 2 * T * cos 30.8 * cos 27

Vertical components of the tension forces = T * sin 30.8
Since there are two ropes, the total force is 2 * T * sin 30.8
Weight = 276 * 9.8 = 2704.8 N
Net downward force = 2704.8 – 2 * T * sin 30.8
Ff = 0.595 * (2704.8 – 2 * T * sin 30.8)
Ff = 1609.356 – 1.19 * T * sin 30.8

Set the two forces equal to each other and solve for T.

2 * T * cos 30.8 * cos 27 = 1609.356 – 1.19 * T * sin 30.8
2 * T * cos 30.8 * cos 27 + 1.19 * T * sin 30.8 = 1609.356
T * (2 * cos 30.8 * cos 27 + 1.19 * sin 30.8) = 1609.356
T = 752.0324 N

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

2 * T * cos 30.8 * cos 27 = 2 * 752.0324 * cos 30.8 * cos 27 = 1151.12

The only difference is the coefficient of friction.

Ff = 0.403 * (2704.8 – 2 * T * sin 30.8)
Ff = 0.403 * (2704.8 – 2 * 752.0324 * sin 30.8) = 779.665 N

Net force = 1151.12 N – 779.665 N = 4371.454 N
To determine the acceleration, divide by the mass.

a = 371.454 ÷ 276
This is approximately 1.3458 m/s^2.

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