1 a 3kg block is at rest on a table the static friction...

1. A 3kg block is at rest on a table. The static friction coefficient between the box and the table is 0.40 and the kinetic friction coefficient is 0.10. A 10N horizontal force is applied to the box. What is the box's acceleration? Show all work and draw the free body diagram to receive full credit. 2. We have a 10kg box that we let slide down a ramp that is angled at 40 degrees to the horizontal. The block has an acceleration of 3.0m/s^2. What is the coefficient of kinetic friction between the box and the ramp? Draw a free body diagram and show all work to receive full credit. (Hint: Net force in the x-direction is ma=mgsin(theta)-F_k … plug in known values to this equation to solve for F_k. Proceed to solve for the normal force and finally calculate the coefficient of kinetic friction.)

Transcript

00:01 For a problem two, you're considering a person dragging and packing crate across a horizontal floor.

00:10 We want to know what is the angle above the horizontal.

00:13 This person should be pulling in order for them to achieve the maximum acceleration.

00:19 So let's draw a picture.

00:21 Here's our floor.

00:23 Here is our crate.

00:24 It has a mass of m.

00:26 And there is friction between the crate.

00:30 And the floor.

00:31 The other things that we know are that it is being pulled with some force f that is at an angle theta above the horizontal.

00:41 And because there's friction, there will be a force of friction that is in opposition to that motion.

00:50 Now, in order to figure out when will we see the maximum acceleration, we're going to need to use newton's second law.

00:58 So we're going to draw another diagram here we'll draw the free body diagram.

01:04 I always start my free body diagrams with the gravitational force because it's an easy one to forget.

01:12 Our box is or the crate rather is on the floor so there will be a normal force directed upwards.

01:22 We have the applied force f off to the angle theta here and then the frictional force off to the left.

01:36 And of course we'll want to find the x and y components of the applied force.

01:44 So the x component will be f cosine theta and the y component will be f sine theta.

01:53 So this crate is going to be accelerating in the horizontal direction.

01:58 So i'm going to apply newton's second law in the horizontal direction in order to try and solve here.

02:08 So in the horizontal direction, the forces that we need to consider are the applied force f cosine theta in the positive x direction and the frictional force in the negative x direction.

02:26 Of course, we need to then find what the frictional force is going to be.

02:30 And for kinetic friction, that's always going to be the coefficient of kinetic friction times the normal force.

02:39 The normal force in this case is going to be determined using newton's second law in the y direction.

02:47 I'm going to assume that you know how to do that.

02:50 But essentially, it must balance the other forces in that direction.

02:56 And so in this case, that is going to be the force of gravity.

03:02 As well as the y component of the applied force.

03:07 And they're in opposite directions, so they're going to be subtracted from each other.

03:15 Okay, now we wanna try and simplify and figure out how can we maximize that acceleration.

03:21 So by doing that, we see math times acceleration is going to be equal to f cosine theta plus sine, theta minus mu kmg.

03:40 And so the only thing that we can change here is the angle theta.

03:48 And so in order to maximize the acceleration, we need to maximize this cosine theta plus sine theta.

03:56 We want this to be a max in order to maximize acceleration.

04:11 We also need to keep in mind that presumably our person wants to accelerate to the right...

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