A uniform stationary ladder of length l and mass m leans...

A uniform stationary ladder of length L and mass M leans against a smooth vertical wall, while its bottom legs rest on a rough horizontal floor. The coefficient of static friction between floor and ladder is μ. The ladder makes an angle θ with respect to the floor. A painter of weight 1/2M stands on the ladder a distance d from its base. Find an expression for the magnitude of the normal force N exerted by the floor on the ladder. Find an expression for the magnitude of the normal force NW exerted by the wall on the ladder. Find an expression for the largest value of dmax for which the ladder does not slip. What is the largest value for d, in centimeters, such that the ladder will not slip? Assume that ladder is 4.5 m long, the coefficient of friction is 0.59, the ladder is at an angle of 43.5°, and the ladder has a mass of 85 kg.

Transcript

00:01 Okay, so we have this ladder that is up against this friction, this wall, and there's a friction force down here.

00:08 So it's on a rough surface down here, and we got our painter here, who is some distance d -u -a from the bottom of the ladder.

00:15 The ladder itself has a length of l and a mass of m, and the painter's mass is half that of the lack.

00:23 So the first thing asks for the normal force from the floor.

00:26 And so we can just look at the net force on this ladder to find the normal force on the floor because there's no friction force here.

00:35 This is a frictionless wall, which means there's no vertical component to the force here, only a horizontal component, the normal force here.

00:45 So the forces acting in the wide direction are just the weight of the ladder, the weight of the painter, and then the normal force up.

00:54 So we can plug those in.

00:56 The net force is going to be zero if we want this ladder to stay stationary.

01:00 So the normal force up has to balance the weight force down of the ladder and the weight force down of the man, the painter.

01:13 So if we plug in the mass of the painter is half the mass of the ladder.

01:18 So we're going to have m over 2 times g, which is we can combine these.

01:23 This becomes three halves m .g.

01:25 That's going to be the normal force from the floor.

01:29 Next, we're looking for the normal force from the wall.

01:33 And for this, we can use the sum of the torques around the point of rotation down here where it's going to be touching the floor.

01:42 I chose this one because that's where all my distances are measured from.

01:47 So that's where it's most useful for me.

01:52 So some of the torques around that point will have to be zero in order for this to not move.

01:59 So we've got two clockwise torques and one counterclockwise torque.

02:03 And i'm going to choose clockwise to be positive in this case, just because that's how i wrote it out, and i didn't want to go back and fix it.

02:12 So clockwise is going to be positive, counterclockwise will be negative.

02:16 So i've got my clockwise torque due to the mass of the ladder, which is going to be the center of mass of the ladder is l over 2.

02:25 So that's the distance.

02:26 That's r and then the weight of the ladder we're looking for the perpendicular part of that weight so m g times the cosine of theta for that force so l over 2 m g cosine theta for the latter and then we can do the same thing for the man where it's d he's some distance d up from the ladder mg cosine sine theta, where this is his mass.

02:54 So this will become m over 2...

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