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

Question

A heavy sled is being pulled by two people. The coefficient of static friction between the sled and the ground is s = 0.587, and the kinetic friction coefficient is k = 0.411. The combined mass of the sled and its load is m = 261 kg. The ropes are separated by an angle = 23°, and they make an angle = 30.4° 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

Net force along vertical direction is zero

N + Tsin(30.4)+Tsin(53.4) = mg = 261*9.8

N = 2557.8 - T sin(30.4)- Tsin(53.4) .....(1)

Net force along horizontal direction is zero

Fs = T cos(30.4) +Tcos(53.4)

sN =  T cos(30.4) +Tcos(53.4)

From (1)

0.587(2557.8 - T sin(30.4)- Tsin(53.4)) =  T cos(30.4) +Tcos(53.4)

By solving above equation we get

Tension in the rope = 674.2 N

If the sled is accelerating

From Newton seocnd law net force along horizontal direction is ma

T cos(30.4) +Tcos(53.4) - kN = ma

T cos(30.4) +Tcos(53.4) -0.411(2557.8 - T sin(30.4)- Tsin(53.4)) = 261*a

(674.2*cos(30.4)) +(674.2*cos(53.4)) -0.411(2557.8 - (674.2 sin(30.4))- (674.2*sin(53.4)))  = 261*a

By solving above equation we get

a =1.13 m/s2

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