A loaded penguin sled weighing 70 N rests on a plane inclined at 20° to the hori
ID: 2043427 • Letter: A
Question
A loaded penguin sled weighing 70 N rests on a plane inclined at 20° to the horizontal. Between the sled and the plane the coefficient of static friction is 0.25, and the coefficient of kinetic friction is 0.19.(a) What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane?
N
(b) What is the minimum magnitude F that will start the sled moving up the plane?
N
(c) What value of F is required to move the block up the plane at constant velocity?
N
Explanation / Answer
The most important part of this problem is drawing a force diagram, labeling all acting forces and choosing your axis. The acting forces are the force of gravity (70N) and the frictional force between the sled and the plane. I would make the x axis parallel to the plane and the y axis would naturally be perpendicular. This means the force of friction is parallel to the x axis but the force of gravity must be broken into its x and y components. Fgx = 7sin20 and Fgy=70cos20. The force of friction can be calculated by multiplying the coefficient of friction by the normal force (note friction is different based on whether the sled is moving or not. a)If the sled is slinding then the net force in the x direction must be negative meaning the x component of gravity is greater than the force of friction. To make it stop sliding the net force must be 0 so subtract friction from Fgx and that will be the amount of force needed to stop the sliding. b)to make the sled move up hill the net force in the x direction must be positive so the required force must be greater than Fgx c)if the sled is moving at constant velocity this means acceleration is 0, meaning net force is 0. Therefore the required force for constant velocity is the same as Fgx.
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