A 30 kg box is at rest on a horizontal floor with friction...

A 30-kg box is at rest on a horizontal floor with friction coefficients Î¼s = 0.65 (static) and Î¼k = 0.55 (kinetic). You push on the box HORIZONTALLY (parallel to the floor) with varying applied force. For each applied force below, find the magnitude of the friction force exerted by the floor on the box (in Newtons), and state whether the friction is static or kinetic. (a) Applied force 100 N. (b) Applied force 150 N. (c) Applied force 190 N. (d) Applied force 200 N. (e) Applied force 250 N. (f) Use these results to draw a continuous graph of the friction force (vertical axis) vs. applied force (horizontal axis), and indicate any "special values" on the axes where the graph changes or does something interesting.

Transcript

00:01 So we have a 30 kilogram box that just rests on a horizontal floor.

00:05 So something like this, that's our floor.

00:08 This is our box m.

00:10 And we have these two coefficients of friction.

00:13 For static friction, we have mu -s -as -equal to 0 .65.

00:18 And for kinetic friction, we have mu -k is equal to 0 .55.

00:25 We push on the box horizontally with a varying applied force.

00:30 So we'll call that force push.

00:35 We know that there's some friction force acting against the direction of the push, and we have normal force acting up from the floor on the box, and we have weight force from the earth pulling on the box.

00:52 We know that in the y direction, we have no acceleration, so we know that net forces in the y direction equals zero.

01:03 This allows us to show that normal force is equal to weight, which is equal to m -g.

01:10 And before i forget, we were told in the beginning that mass is equal to 30 kilograms.

01:17 We also know that in general, force of friction is equal to either we have static friction.

01:25 It will be equal to mu -s times n.

01:28 This is actually the maximum value for static friction.

01:30 We'll get that to that in a second.

01:31 And the value for kinetic friction is equal to u k times n.

01:39 So we know that n is equal to 30 kilograms times 9 .8 meters per second squared.

01:53 This gives us a normal force value of 294 newtons.

02:00 So for our max value of static friction, we have 0 .6 .6.

02:09 Times 294 newtons and for kinetic friction we have 0 .55 times 294 newtons and we get that our max value for static friction is equal to 191 .1 newtons and our value for kinetic friction is equal to 191 .1 newtons and our value for kinetic friction is equal to 161 .7 newtons.

02:43 This will come in handy.

02:46 And now we want to see what happens as we vary that applied force, force push.

02:53 So the first applied force, we have force of push is equal to 100 newtons.

02:59 This is going to be less than the force of static friction at its maximum value.

03:11 So this implies that the box remains in motion.

03:17 And that whatever force of friction we have here molds itself to be equal to force push.

03:26 Sorry, not that it remains in motion, it remains stationary.

03:38 So here, force of friction molds itself to the applied force such that we remain stationary.

03:44 And so here we know that force of friction is equal to 100 newtons.

03:50 For b, we're told that the applied force is now 150 newtons.

03:59 And similarly to above, this applied force is less than the maximum value for static friction.

04:09 We are again stationary.

04:14 And this implies that force of friction here is equal to the applied force, which is equal to 150 newtons...

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