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

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=0.595 , and the kinetic friction coefficient is k=0.467 . The combined mass of the sled and its load is =336 kg . The ropes are separated by an angle =27° , and they make an angle =30.8° with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving?

To help with solving:

There are four forces acting on the sled: gravity pulls downward, the normal force from the ground pushes upward, friction pushes away from the people, and the combined force of the ropes pulls upward at an angle to the horizontal. Draw a freebody diagram and use Newton's laws to determine the mathematical relations between these forces. To decide which laws to use, consider the sled's vertical and horizontal acceleration components. You will also need to know the formula for the maximum static friction force s,max s,max=s where s is the static friction coefficient and is the normal force.

Explanation / Answer

Horizontal component of tension = T cos 30.8 * cos 27 .

Since, there are two ropes. so,

Total Horizontal component of tension,

Tx = 2T cos 30.8 * cos 27 ......(1)

Vertical component,

Ty = 2T sin 30.8

Net vertical force = mg -  2T sin 30.8

N = 336*9.8 - 1.024*T

Static Friction force, f = uN

f = 0.595 ( 336*9.8 - 1.024*T)

f = 1959.21 - 0.6093 T .......(2)

From eq(1) and (2),

2T cos 30.8 * cos 27 =  1959.21 - 0.6093 T

1.5306T + 0.6093 T = 1959.21

T = 915.5 N

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