2. A uniform ladder of length L rests against a smooth...

2. A uniform ladder of length L rests against a smooth vertical wall as shown below. The mass of the ladder is m and the center of mass of the ladder is located at the mid-point of the ladder. There is no frictional force between the ladder and the wall, but there is friction between the base of the ladder and the floor with the coefficient of static friction given by ?. The ladder is placed against the wall such that the angle that the ladder makes with the floor is 56°. (a) Identify the forces and torques acting on the ladder. (b) What minimum value for the coefficient of static friction ? is needed to keep the ladder in equilibrium? [NOTE: It may look like you don't have enough information to answer this question (like masses, lengths, etc.) but, trust me, you do. If you do the algebra all of those unknowns end up canceling out.]

Transcript

00:02 All right, so i just went ahead and drew a diagram.

00:08 So right here i have labeled the orientation of my axis.

00:14 So up is positive on the y -axis.

00:17 Right is positive on the x -axis.

00:19 I guess i can write this like this.

00:21 And then rotationally, counterclockwise size is positive.

00:25 So that's what that means.

00:28 And then also i drew out all of our forces.

00:32 And i'll explain them real quick.

00:33 So we have our ladder and we know that its weight, mass, it's mass, time gravity is pointing straight down.

00:45 It's at an angle theta equals 56 degrees with the ground.

00:53 And so there's also friction between the bottom of the ladder and the ground.

01:00 So we can get our frictional force.

01:06 And it's pointing keeping the ladder from sliding to the left so it's pointing to the right and then also there's a normal force from the floor up on the ladder pushing against gravity and then also there is a force from the wall acting on the ladder keeping it from going through the wall so the normal force acts to the left all right so first thing we can do.

01:39 Do is write out all of our equations.

01:40 That's part a.

01:41 We need to write out all of the forces and torques on it on the ladder.

01:48 So i'm going to go ahead and write using newton's summation law notation.

01:53 So the first forces we'll think about are the forces in the y direction.

02:02 So because the ladder is assumed to be static, not moving, we know that some of the forces in the the wide direction will be equal to zero.

02:12 And we, according to our convention, we can write the f2 is minus the weight of the ladder.

02:24 That's going to be equal to zero, because those are the forces in the y direction.

02:28 Likewise, for the x direction, we have the frictional force.

02:34 And the sum of the force in the next direction are also going to be zero.

02:37 So we've got the frictional force.

02:40 And then we subtract the normal force from the wall onto the ladder from that.

02:47 And that will be our forces in the x direction.

02:50 All right.

02:51 So those are all of our forces.

02:53 Now we want to look at our torques.

02:57 So i'll write the sum of torques.

03:00 And to make the problem easier on us, i went ahead and set the center around which we're considering our torques to be at the base of the ladder.

03:09 Because there are two forces at this point on the ladder, if we set our torque center there, then all the torques from these two forces will be equal to zero because you have to multiply the force times the distance from the point you're measuring the torques around in order to get your torque.

03:28 So these torques will go to zero.

03:31 Zero for the torque.

03:34 All right.

03:37 Let's cluttered.

03:38 All right.

03:39 So we know that the size, some of our torques are going to be equal to zero because, again, the latter is static.

03:50 So let's look at what torques we have.

03:54 So again, these forces will not apply torque on about this point, but we have two forces that will.

04:01 We've got the weight and then the normal force from the wall.

04:04 So let's think about the weight.

04:09 So we have the weight pointing down, but we want the, we want the four, we want, we want, the torque to be perpendicular to, we want to measure the torque perpendicular to the axis that we're measuring torque around.

04:24 So since we're measuring along this axis, we want to get a perpendicular force and measure that perpendicular torque.

04:33 So i went ahead and drew a triangle, and using seam geometry, you can find theta to be this angle right here.

04:42 And then cosine a theta is going to be the adjacent.

04:45 So if you multiply the force times the cosine of theta, and then multiply that times the distance from the point that we're measuring around, which is going to be l divide by 2, we can get that torque.

05:04 And yeah, that's fine.

05:08 All right, so the next one.

05:09 So the next torque is going to be the component perpendicular to the rod, and we find that the perpendicular component of the force is going to be f1 times the sine theta.

05:26 So we're going to subtract it, because it's going the opposite direction of the other torque applied by the weight.

05:35 So we have got f1, sine theta...

Question

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