A loaded penguin sled weighing 88 N rests on a plane inclined at angle ? = 18 So

Question

A loaded penguin sled weighing 88 N rests on a plane inclined at angle ? = 18

Explanation / Answer

The most important part of these mechanic questions is to understand whats going on. You should start by drawing a free-body diagram. On the diagram you should have 3 forces acting on the sled: Fs (for static friction) N (the normal force or force keeping sled from sinking into plane), Fg (gravitational force) = mg. Note what we're given: Weight 88N this is Mass* Gravity, µs=0.28, µsk=0.20, ?=18° and the keywords: (parallel to plane- 90° somewhere, prevent from slipping- at least stationary, minimum magnitude, constant velocity- Fnet = ma = 0.) For part (a) We'll interest ourselves with the given weight mg= 88N, µs = 0.28, and the angle between plane and horizontal ?=18°. One more thing! For (a) we must recall that if the sled is not slipping/moving its net force is zero (by newtons laws). So we will consider Fnet = ma = 0. The next important part is to make sure you're free body diagram shows the forces in the right directions. This will clarify the situation and make the question much easier to understand. Remember the normal force is perpendicular (at right angle or 90°) to the PLANE/slope the sled is resting ON. Mg acts directly downward (at 90° to HORIZONTAL). Fs acts opposite to the direction of motion or intended slippage etc. You need to choose the plane the sled is on for your X axis and the direction of N for your y axis. Now we start! From F = ma = 0 we'll get for our Y AXIS: N-mgsin? = 0 (The normal force stops the sled from dropping through the slope. It is equal and opposite) N - 88N*sin252 = 0 (We get 249? from measuring the angle between OUR chosen x axis and the direction the weight is acting in - mg) Just rearrange and ---> N = 70N * sin252 Next we know Fs = µsN -----> Fs= 0.22* 70N * sin252 To find the force needed to keep this stationary we must recall that Fnet must equal 0 on both axis! So we must check the X AXIS now. Again from F= ma = 0 We have Fs (static friction) + F(the force we need to find) - weight (on our x axis) = 0 (It is important to resolve the weight into the x and y components for the axis we chose, also the angle should be measured from our x axis). This becomes: 0.28* (88N * sin252) + F - (88*cos252) = 0 Make F subject of equation --> F = (88* cos252) - [0.28 * (88N * sin252).(This is easy with a calculator!) For this question we're trying to find what force we need to help the static frictional force to keep the sled at least stable. If we push any harder the sled will start moving up the plane for (b). For (c) constant velocity is a hint that tells us that the sum of the forces is still 0 even though the sled is moving upward. Again on our x axis we'll have F = ma = 0 but important to NOTE that while the sled is moving the frictional force changes from static friction to kinetic friction. This indicates that we must use Fk instead of Fs. Fk will act in the opposite direction now because the direction of motion changes. Fk = 0.13 * N( the normal force we worked out earlier) F2 - (mgcos252) - 0.20* (88N * sin252) = 0 F2 = (mgcos252) + [0.2 * (88N * sin252) = (....)N

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