A ladder of length l and mass m1 leans against the side of a...

A ladder of length L and mass m1 leans against the side of a building, making an angle θ with the horizontal. Assume the ladder is uniform and the side of the building is frictionless. (a) A person of mass m2 stands on the ladder a distance x along the ladder from the bottom. What are the horizontal and vertical forces the ground exerts on the ladder? (Use any variable or symbol stated above along with the following as necessary: g.) Fhorizontal= Fvertical= (b) If the ladder is just on the verge of slipping when the person is a distance d along the ladder from the bottom, what is the coefficient of static friction between ladder and ground? (Use any variable or symbol stated above along with the following as necessary: g.) μ =

Transcript

00:01 In this question, a ladder is given where the walls are frictionless and a person of mass m2 stands on a ladder at a distance x from the bottom.

00:21 Let's suppose this is a person standing of mass m2.

00:28 We have to find the horizontal and the vertical forces on the ground.

00:36 Also in the question it has been given that the the length of the ladder is l and mass of the ladder is m1.

00:46 So now let's suppose this is x, this is angle alpha, and this mass m1 is at the midway between the length of the ladder.

01:07 Now let's mark the points a, b.

01:11 Now taking moments from point a we can write here in the diagram m1g is acting downwards similarly m2g is acting downwards and the vertical reaction force v is acting upwards at point b so now writing moment of forces taking from point a we can write m1g into this particular perpendicular length now this per length is l by 2 cost theta as one can see let's see if this is o here, in triangle o, a, b, this is length l, that is length of the ladder, and this is alpha, then this ob is equals to l cross alpha, and half of this length, that is from here to this point, will be equal to ob by 2, that is l by 2, cos alpha.

02:31 Similarly we can write here that this particular length equals to m1g into l by 2 cos alpha plus m2g into l minus one can write here l cos alpha minus x cos alpha will be equal to ob b and x cos alpha will be this particular length so we have to basically calculate this length so we have taken the difference of it that is l cost alpha minus x cos alpha equals to this vertical force v acting at point b and this total vertical length that is ov which is equals to v into l cost alpha so solving for v we will get v equals to that is vertical force at point b v equals to m1g by 2 plus m2g into 1 minus xy l.

04:20 So this is the vertical force acting on the ladder at point b.

04:26 Similarly, the horizontal force can also be calculated by taking moment at point b.

04:35 Now if it let's say this is a -o -b.

04:41 Now horizontal force, let's say fh acting in this direction, if we take moment from point b and also mark the other forces.

04:52 Here it will be m1g and here it will be m2g.

04:59 Now in this case we have to take basically that is length in the vertical direction.

05:05 So we first mention the length...