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 mu_s = 0.603, and the kinetic friction coefficient is mu_k = 0.403. The combined mass of the sled and its load is m = 291 kg. The ropes are separated by an angle phi = 22 degree, and they make an angle theta = 31.1 degree 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

N = mg

N = 291 * 9.8 = 2851.8 N

static frcition force

fs = mu_s * N

fs = 0.603 * 2851.8 = 1719.6354 N

one person is exerting force = F

force by two people = 2F

each force making angle 31.1 degree from horizontal and 11 degree to forward

Fp = 2*F*cos31.1 * cos11

Fp = 1.681 F

this will be equal to fs for moving

1.681F = 1719.6354

F = 1022.94 N this is also the tension

T = 1022.94 N

part b )

now sled is moving so there will be kinetic friction

fk = mu_k*N

fk = 1149.2754 N

Fnet = fs -fk

ma = fs - fk

a = (fs-fk)/m

a = 1.96 m/s^2

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