Asked by josh
A sled (mass 310 kg) is pulled across a stone floor with a coefficient of kinetic friction of 0.9. The rope that is used to pull it is at an angle alpha of 55 degrees with the horizontal. How hard (i.e., with what magnitude force) do you need to pull to make the sled speed up with an acceleration of 6.33 m/s2?
Answers
Answered by
Damon
T = tension in rope
rope horizontal force = T cos 55
= .574 T
rope force up = T sin 55
= .819 T
weight = m g = 310(9.81) = 3041 N
normal force on ground = weight - rope force up
= 3041 - .819 T
friction force max = .9(3041-.819 T)
F = m a
.574 T - .9(3041-.819 T) = 310(6.33)
solve for T
rope horizontal force = T cos 55
= .574 T
rope force up = T sin 55
= .819 T
weight = m g = 310(9.81) = 3041 N
normal force on ground = weight - rope force up
= 3041 - .819 T
friction force max = .9(3041-.819 T)
F = m a
.574 T - .9(3041-.819 T) = 310(6.33)
solve for T
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