1 1140 n crate is being pushed across a level floor at a...

1) A 1,140-N crate is being pushed across a level floor at a constant speed by a force F of 370 N at an angle of 20.0° below the horizontal. (a) What is the coefficient of kinetic friction between the crate and the floor? (Enter your answer to at least three decimal places.) (b) If the 370-N force is instead pulling the crate at an angle of 20.0° above the horizontal, as shown in the figure b, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).

Transcript

00:01 All right, so we have a box being pushed across a horizontal floor at a constant speed, and we're told the weight of the box is 1 ,140 newtons, and it's pushed across with a force of magnitude 370 newtons that is directed 20 degrees below the horizontal.

00:18 And so we want to know what's the magnitude of the friction, or the coefficient of friction i mean, in this part.

00:24 So if we look at a free body diagram, we're going to have the weight of the box going down.

00:30 The normal force going up, which is not equal to the weight necessarily, because we have a force that is, um, has components of 370 newtons times the cosine of 20 going horizontally.

00:44 And then it has a magnitude of 370 newtons times the sign of 20 going upwards.

00:52 And then we have the frictional force opposing this motion.

00:56 Hopefully that's not the, the most confusing diagram reversing.

00:59 But so what we have then is if this is moving at constant speed, the frictional force is equal in magnitude to 370 newtons times the cosine of 20.

01:11 And so that comes out to 347 .69 neutens.

01:20 But the normal force is a little more subtle.

01:23 So the normal force, if we look at this diagram, is going to be the weight minus 370 newtons times.

01:30 The sign of 20 because that force since it's directed at an angle part of it lifts the box upwards and so it reduces the normal force and so this is going to be like 1 ,140 newtons minus 370 times the sign of 20 and so this gives us a normal force of let me just go ahead and write the final result 1 ,113 .4 .5 newtons so it's the code the coefficient of friction times this normal force is equal to the frictional force that we found earlier.

02:07 So our coefficient of friction is going to be 347 .69 newtons divided by 1 ,113 .45 newtons.

02:20 And so we do that, we get something like 0 .343.

02:28 That's our coefficient of kinetic friction here.

02:30 Part b asks if the force is now pulling the block at an angle above the 20 above the horizontal, what will be the acceleration of the crate? so now our free body diagram looks a little different...

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