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 s = 0.635, and the kinetic friction coefficient is k = 0.435. The combined mass of the sled and its load is m = 276 kg. The ropes are separated by an angle = 29°, and they make an angle = 32.4° 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

We use the static coefficient, 0.635
The normal force holding the sled to the ground is mg = 276(9.8)
The normal force is diminished by the vertical component of the pull.
2Fcos(29/2)sin32.4
The horizontal component of the pull must overcome the static friction coefficient times the normal force:
2Fcos(29/2)cos32.4 = mg - 2Fcos(29/2)sin32.4
F(2cos(29/2)cos32.4 + 2cos(29/2)sin32.4) = mg
F = mg/(2cos(29/2)cos32.4 + 2cos(29/2)sin32.4)
F = 276(9.8)/(2cos(29/2)cos32.4 + 2cos(29/2)sin32.4) = 2211.9525Kg(m)/s^2

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