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 -0.627, and the kinetic friction coefficient is = 0.395. The combined mass of the sled and its load is m = 381 kg. The ropes are separated by an angle = 28, and they make an angle = 32.1° with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving? Number 1596.9145 N If this rope tension is maintained after the sled starts moving, what is the sled's acceleration? Number 2.32 m/ S

Explanation / Answer

According to the given problem,

a)The force that causes the sled is equal to the sum of the horizontal components of the tension forces. The force that causes the sled is the friction force. There are two forces that affect the friction force. The vertical components of the tension forces are the upward forces. The total weight is the downward force. 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 32.1o
To determine the component of these forces that are parallel to the direction the sled is moving, multiply be cos 28o.

T * cos 32.1o * cos 28o
Since there are two ropes, the total force is 2 * T * cos 32.1 * cos 28

Vertical components of the tension forces = T * sin 32.1
Since there are two ropes, the total force is 2 * T * sin 32.1
Weight = 381 * 9.8 = 3737.6 N
Net downward force = 3737.6 – 2 * T * sin 32.1
Ff = 0.627 * (3737.6 – 2 * T * sin 32.1)
Ff = 2343.5 – 1.254 * T * sin 32.1

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

2 * T * cos 32.1 * cos 28 = 2343.5 – 1.254 * T * sin 32.1
T = 2343.5 ÷ 2.1623 = 1083.8 N
This is approximately 1084 N.

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

Ff = 1621.3N

The only difference is the coefficient of friction.

Ff = 0.627 * (3737.6 – 2 * T * sin 32.1)
Ff = 0.395 * (3737.6 – 2 * 1084 * sin 32.1) = 1021.3 N
This is approximately 1021.3 N.

Net force = 1621.3 - 1021.3 =600N
To determine the acceleration, divide by the mass.

a = 600 ÷ 381
This is approximately 1.575 m/s2.

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