Question

A heavy sled is being pulled by two people. The coefficient of static friction between the sled and the ground is μs = 0.595 and the kinetic friction coefficient is μk = 0.403. The combined mas of the sled and its load is m =276 kg. The ropes are separated by an angle φ = 27 degrees, and they make and angle of θ = 30.8 degrees with the horizontal. assuming both ropes pull equally hard, a) what is the minimum rope tension required to get the sled moving? and b) if the tension is maintained after the sled starts moving, what is the sled's acceleration?

Answer 1

(a) The tension in each rope is, T2T each rop COS 2 net 270 2T, COs each rope 2 1.941each rope Apply Newton's second law in the vertical direction as follows: ΣΗ. ma net T sin 0 N= mg net N mg-T sin e net sin (30.80 (276 kg) (9.8 m/s2)-1.947e each rope -2704.8-0.99T, each rope The force on the sled is, F N (0.595) (2704.8-0.99T, each rope 1609.36-0.59T each rope Apply Newton's second law in the horizontal direction as follows: T cos F net cos (30.8°) 1609.36-0.59T, each rope (1.947 each rope 1.67T 1609.36-0.59T each rope each rope Tach rore 713.25 N

Therefore, the minimum tension in the rope to get the sled moving is 713.25 N (b) The frictional force when the sled starts moving is (0.403) (1609.36-0.59T, each rope 648.57-0.24(713.25 N) -478.94 N Apply Newton's second law in the horizontal direction as follows: = ma net T cos- F ma net T. cos - F net each rope COS(30.8°)-478.94 N 276 kg 1.94T 1.94(713.25 N) cos (30.8)-478.94 N 276 kg 2.57 m/s