Asked by Ameer
A child pulls on a rope of of sled that is stuck in snow at an angle of 34° with respect to the horizonal. The mass of the sled is 15 kg. Determine the applied force needed to get the sled moving if the coefficient of static friction between the sled and snow is 0.12 [Hint: when caculatING the force of friction, there are two forces upward]
Answers
Answered by
Henry2
M*g = 15 * 9.8 = 147 N. = Wt. of sled.
Fn = 147 - F*sin34 = 147-0.56F = Normal force.
Fs = u*Fn = 0.12(147 - 0.56F) = 17.64 - 0.067F = Force of static friction.
F*Cos34 - Fs = M*a.
0.83F - (17.64-0.067F) = 15 * 0,
0.83F + 0.067F -17.64 = 0,
0.9F = 17.64,
F = 19.7 N. = Applied force.
Fn = 147 - F*sin34 = 147-0.56F = Normal force.
Fs = u*Fn = 0.12(147 - 0.56F) = 17.64 - 0.067F = Force of static friction.
F*Cos34 - Fs = M*a.
0.83F - (17.64-0.067F) = 15 * 0,
0.83F + 0.067F -17.64 = 0,
0.9F = 17.64,
F = 19.7 N. = Applied force.
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