The diagram below is a top-down view of two children pulling a 11.1 -kg sled alo
ID: 2130362 • Letter: T
Question
The diagram below is a top-down view of two children pulling a 11.1-kg sled along the snow. The first child exerts a force of F1 = 14 N at an angle ?1 = 45
The diagram below is a top-down view of two children pulling a 11.1-kg sled along the snow. The first child exerts a force of F1 = 14 N at an angle ?1 = 45 degree counterclockwise from the positive xdirection. The second child exerts a force of F2 = 8 N at an angle ?2 = 30 degree clockwise from the positive x direction. Find the magnitude and direction of the friction force acting on the sled if it moves with constant velocity. What is the coefficient of kinetic friction between the sled and the ground? What is the magnitude of the acceleration of the sled if F1 is doubled and F2 is halved in magnitude?Explanation / Answer
a) Fnet = 12cos45i + 12sin45 j + 8cos30 i - 8sin30 j
= 15.41 i + 4.49 j N
so friction will be same as Fnet and in opposite direction .
say f = - 15.41i - 4.49 j N
magnityde = sqrt(15.41^2 + 4.49^2) = 16.05 N
direction = 180 + tan-1(4.49/15.41) = 196.25 degrees
b) f =umg
16.05 = u x 11.3g
u = 0.145
c) that means F = 16.05 x 2 and f = 16.05/2
then F- f = 16.05 x 3/2 = 11.3a
a = 2.13 m/s2