The diagram below is a top-down view of two children pulling a 11.3-kg sled alon

Question

The diagram below is a top-down view of two children pulling a 11.3-kg sled along the snow. The first child exerts a force of F1 = 12 N at an angle ?1 = 45

The diagram below is a top-down view of two children pulling a 11.3-kg sled along the snow. The first child exerts a force of F1 = 12 N at an angle ?1 = 45 degree counterclockwise from the positive x direction. 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? Doubling or halving a force will not necessarily have the same effect on the resultant. m/s2

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

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