A uniform beam AB has length 6m and mass 45 kg. One end of a...

A uniform beam AB has length 6m and mass 45 kg. One end of a light inextensible rope is attached to the beam at the point 2.5 m from A. The other end of the rope is attached to a fixed point P on a vertical wall. The beam is in equilibrium with A in contact with the wall at a point 5 m below P as shown in Figure Q4. The rope is taut and at right angles to AB. Use g = 9.81 m/s". P 5 m 3.5 m 2.5 m A B Figure Q4 (a) Draw the force diagram, indicating all the forces acting on the beam AB. (b) Using vectors, write down the equations for all the forces acting on the beam AB and its position vectors from A. Apply equilibrium principles and take torques about A. (c) Solve the magnitude of the tension, |T| in the rope. (d) Solve the magnitude of the horizontal component, |FH| and the vertical component, |FV| of the force exerted by the wall on the beam at A.

Transcript

00:01 High, so we are given that a uniform beam of length ab is attached to the wall in such a way that you can see into the figure, whose mass is m is equal to 45 kg and length of this beam lab you will say 6 meter.

00:31 One end of the light extendable rope is attached to the beam at an end of 2 .5 meter from it.

00:39 So from a point up to 2 .5 meters and a string is connected throughout this support at a p point by this beam, makes an angle of 90 degree with this, obviously perpendicularly it is connected.

00:57 So the beam is in equilibrium position with a in contact and from a point to this p point the distance is 5 meters.

01:05 So in a part we have to draw the force diagram indicating all the forces acting on the beam ab.

01:13 So let us get started with solution.

01:17 So first of all for the a part as we have to draw the free body diagram.

01:24 So let us draw this.

01:26 So first of all i will say this is what the free support, this i am considering the beam and this is what the string here, likewise which is up to distance of 2 .5 meters.

01:41 This is about 5 meters and this complete about 6 meters.

01:46 So basically we need to find the center of mass.

01:49 So obviously it is going to be at the center of its, so here will be its center of mass at which the downward force will be exerted, which will be basically the weight of its w which is equal to mass into gravity and this is at 3 .5 meters from a.

02:08 This is the point p.

02:10 Now obviously in the string there will be a tension in a 4 direction.

02:15 This is the p point, this is a point.

02:18 So obviously it will make some angle with respect to the vertical, let's say theta.

02:24 At a point some reactions will be there.

02:29 So let me show you it's by another color.

02:33 So one will be ry of a and one is rx of a and here we can draw the vertical component of it with respect to theta.

02:44 Likewise this because tension will also have the two components.

02:48 So this is what the answer for a part.

02:51 Now let's go throughout the b part in which we have to find by using the vectors we have to write down the equation for all the forces acting on the beam about point a.

03:04 Then we have to solve the magnitude of t.

03:06 So first of all from the figure we can apply sine theta property.

03:11 This should be equals to opposite upon abtaneous that is 2 .5 by 5.

03:15 This will comes out equals to 1 by 2.

03:16 That means theta should be equals to sine inverse of 1 by 2 is 30 degree.

03:22 So for vertical equilibrium, for vertical equilibrium let us apply summation of force fv is equals to 0 according to the newton's law...

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