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 μs = 0.571, and the kinetic friction coefficient is μk = 0.395. The combined mass of the sled and its load is m = 306 kg. The ropes are separated by an angle φ = 23°, and they make an angle θ = 29.7° 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?

Answer 1

The force of gravity on the sled is:

Fg = m g

Fg = 306 * 9.8

Fg = 2999.8 N

The force of static friction is:

Fsf = us * Fg

Fsf = 0.571* 2999.8

Fsf = 1712.3 N

Say the force exerted by ONE person is F.

The two persons together exert 2F to get the sled moving.

Each of those forces make an angle of 29.7degrees to horizontal and 23/2 = 11.5 degrees from forward.

So the component of 2F in the direction of the motion is:

Fm = 2 F cos(29.7) cos(11.5)

Fm = 0.8512 F

That force must be the same as Fsf to get the sled moving, so

0.8512 F = 1712.3

F = 2011.6 N

That is also the tension in each rope.

T = 2011.6 N < - - - - ANSWER

Once the sled is moving, the friction 'switches' to 'kinectic-mode'

The kinetic friction is:

Fkf = uk* Fg

Fkf = 0.395 *2999.8

Fkf = 1184.9 N

The net force exerted on the sled now is:

Fn = Fsf - Fks

Fn = 1712.3 - 1184.9

Fn = 527.4 N

The acceleration of the sled is:

F = m a

527.4 = 306 * a

a = 1.723 m/s < - - - - ANSWER