To move a large crate across a rough floor, you push down on...

Name: To move a large crate across a rough floor, you push down on it at an angle of 21°, as shown in Figure 6-16. Find the force necessary to start the crate moving, given that the crate is m = 34 kg and the coefficient of static friction between the crate and the floor is 0.62.
Uploaded: 2021-12-31T16:05:59-08:00
Duration: 9 min 33 s
Channel: Linda Winkler
Description: To move a large crate across a rough floor, you push down on it at an angle of 21°, as shown in Figure 6-16. Find the force necessary to start the crate moving, given that the crate is m = 34 kg and the coefficient of static friction between the crate and the floor is 0.62.

To move a large crate across a rough floor, you push down on it at an angle of 21°, as shown in Figure 6-16. Find the force necessary to start the crate moving, given that the crate is m = 34 kg and the coefficient of static friction between the crate and the floor is 0.62.

Transcript

00:01 So a quick reminder about how static friction works.

00:04 If you push on an object on a surface with static friction, you can push harder and harder and harder and harder, and the object may not move.

00:16 The reason being static friction acts kind of like a glue in between the surface of the floor, for example.

00:25 And in this case, we'll think of a box.

00:27 So you can continue to push and push and push until you break the glue.

00:34 And so the way we talk about the force of static friction, little f sub s, it is less than or equal to the coefficient mu sub s times the normal force, which is pushing the two surfaces together.

00:51 So remember when you glue objects together, you have to get the bonding to work by pushing the surfaces together.

00:59 And that is called the normal force.

01:03 Now the critical condition beyond which the box will become unglued, we'll call that critical condition.

01:12 It doesn't mean anybody's getting in bad shape.

01:15 It just means that the static friction has reached its maximum value.

01:22 That is equal to mu -s, times the normal force.

01:27 So as an example of this, we're going to figure out the force needed to get a 34 -kilogram crate to move on a surface with a given static friction coefficient.

01:40 And as usual, what we want to think through with forces is newton's second law.

01:50 So i usually like to draw a force diagram, a free -body diagram, when i'm doing these kind of things.

01:57 But right at the critical point, what we know is we have the critical condition.

02:04 And we also have the sum of the forces in the x direction just barely equal to zero.

02:11 And of course, the sum of the y forces is also equal to zero being on a surface.

02:19 Looking at a free body diagram, we of course have the weight downwards.

02:25 M .g is 34 kilograms times 9 .8 meters per second squared.

02:35 Newton's, that n is for newton's.

02:44 So that's about 33.

02:47 Newton's.

02:48 I'll go ahead and calculate that.

02:50 The force downwards has both an x and a y component, being at 21 degrees.

03:03 The static friction is opposite along the surface, and there is the normal force.

03:12 I'll call it f sub -n, just so we don't get it mixed up with newton's.

03:25 That's always a little awkward to have more than one meaning for a symbol.

03:30 So let's divide our motion up into two different directions...

Question

To move a large crate across a rough floor, you push down on it at an angle of 21°, as shown in Figure 6-16. Find the force necessary to start the crate moving, given that the crate is m = 34 kg and the coefficient of static friction between the crate and the floor is 0.62.

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