Asked by Anonymous
The Grizwald’s have just chopped down a Christmas tree and have decided to pull it out of the woods on a sled over the snow. Clark has attached a rope to the sled and is pulling the tree/sled system (mass = 22 kg) with a force of 50.0 N at an angle of 32˚ above the horizontal. The tree/sled is moving at constant speed. (a) Draw a force diagram of the situation. (b) Create qualitative vertical and horizontal equations for this situation (one each). (c) What is the coefficient of kinetic friction between the snow and the sled?
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Answered by
Nonetheless
Force diagram:
Fn pointing up
Fg pointing down
Fa pointing right/upwards at 32 deg. to horizontal
Fk pointing left
Vertical forces equation (up is positive):
Sled is not moving up or down so:
Fn + Fg + Fa = 0
Fn + (-9.81)(22) + 50 sin 32 = 0
Use the above to solve for Fn.
Horizontal forces equation:
Horizontal velocity is constant so:
Fa + Fk =0
50cos32 + (-uk)(Fn) =0
Plug in Fn that you found above and solve for uk, the coefficient of kinetic friction.
Fn pointing up
Fg pointing down
Fa pointing right/upwards at 32 deg. to horizontal
Fk pointing left
Vertical forces equation (up is positive):
Sled is not moving up or down so:
Fn + Fg + Fa = 0
Fn + (-9.81)(22) + 50 sin 32 = 0
Use the above to solve for Fn.
Horizontal forces equation:
Horizontal velocity is constant so:
Fa + Fk =0
50cos32 + (-uk)(Fn) =0
Plug in Fn that you found above and solve for uk, the coefficient of kinetic friction.
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