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 = 0.603, and the kinetic friction coefficient is ?k = 0.411. The combined mass of the sled and its load is m = 336 kg. The ropes are separated by an angle ? = 21°, and they make an angle ? = 31.1° 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

let the tension required to make the sled start moving is T. (in each rope)

then first we have to find their component acting on the same plane as the sled is and then acting along the direction of movement of the sled.

so first component in the plane of the slade=T*cos(theta)=0.856*T

then component of this force along the direction of motion=0.856*T*cos(21/2)=0.84193*T

similarly, the other rope will exert 0.84193*T N along the direction of motion.

net force=2*0.84193*T=1.6839*T

this force is required to overcome the static friction exerted on the sled.

static friction=static friction coefficient * normal force

here normal force=weight of the sled

==> static friction=static friction coeffcieinct*weight of the sled

=0.603*336*9.8=1985.6 N

==> 1.6839*T=1985.6

==>T=1179.1 N

hence tension in each rope should be 1179.1 N to get the sled start moving.

part b:

once it started to move, the friction acting on it will change to kinetic friction.

kinetic friction force=kinetic friction coefficient*normal force

=0.411*336*9.8=1353.3 N

then net force acting on the sled=1985.6-1353.3=632.26 N

then acceleration of the sled=force/mass=632.36/336=1.8817 m/s^2

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