The diagram below is a top-down view of two children pulling a 10.6-kg sled alon
ID: 1783235 • Letter: T
Question
The diagram below is a top-down view of two children pulling a 10.6-kg sled along the snow. The first child exerts a force of F1 direction. The second child exerts a force of F2-6 N at an angle 2-30° clockwise from the positive x direction. 11 N at an angle 61 45° counterclockwise from the positivex Fi F2 (a) Find the magnitude and direction of the friction force acting on the sled if it moves with constant velocity. magnitude 13.8 direction How is the direction of the resultant related to its components?o counterclockwise from the +x-axis (b) What is the coefficient of kinetic friction between the sled and the ground? 0.133 (c) What is the magnitude of the acceleration of the sled if Fi is doubled and F2 is halved in magnitude? Doubling or halving a force will not necessarily have the same effect on the resultant. m/sExplanation / Answer
(A) moving with constant veloity means a = 0
so Fnet = F1 + F2 + f = 0
f = - 11(cos45i + sin45j) - 6(cos30i - sin30j)
f = - 12.97i - 4.78j N
magnitude = sqrt(12.97^2 + 4.78^2) = 13.8 N
direction = 180 + tan^-1(4.78/12.97)
= 200 deg
(b) uk = f / N = 13.8 / (10.6 x 9.8)
= 0.133
(c) if uk does not change then friction will always be have same value.
f = 13.8 N