00:01
Hello, it is given in the question that the length of the beam l equals 5 meter and the weight of the beam w equals 300 newton and the length at which the person stands l prime that is the length at which the person stands equals 1 .50 meters and the weight of the person prime the weight of the person here is 600 newtons and the theta angle that is the angle of inclination of the cable with respect to the beam equal to 58, 53 degrees.
00:39
Now let us draw a figure to the given question.
00:42
So here this is the wall over here and this is the beam over here, this is the beam over here, this is the beam and this is the cable which is connecting the given beam to the wall.
00:54
And let us consider t is the tension in the given beam, this is the tension in the given cable over here and this tension has two components, one is t sine theta because this is is the angle theta over here the angle of inclination of the cable with respect to the beam and this here this component is t cause theta so this is the two components of the tension t now the reactionary force that is the reactional force of the wall on the beam let us consider that is equal to r and we have to find the value of r in the question so we have to find the value of r in the question reactionary force of the wall on the on the beam so this has two component one is the y component over here sorry this is the x component and this one is the y component over here.
01:36
Now the weight of the beam is acting downwards and that is here is 300 newton and the distance of this distance over here that is equal to 2 .5 meter because the length of the whole beam that is equal to 5 meters length of the beam equal to 5 meters over here.
01:53
So therefore this length the half of the length would be equal to 2 .5 meter that is the length of the center of mass from the wall that is 2 .5 meter and the let us consider the person is standing at this particular point and its weight or her weight is acting downward that is 600 newton is acting downward and this distance over here this distance over here is 1 .50 meters now r is the reaction force of the wall on the beam this r is the reaction force of the wall on the beam now the net torque about a equals 0 so let us consider this is the point a and let us find the torque about point a but the torque about point a is equal to 0 because the system is in equilibrium so since the system system system system is in equilibrium.
02:44
So torque at point a, torque about a, point a equals 0.
02:59
And what is the torque about point a? that is equal to 300 multiplied 2 .5 plus 600 multiplied 1 .5 and that equals t multiplied 5, that is the length of the beam over here multiplied at sine 53.
03:17
So this torque that is the torque due to the weight of the given beam is acting in the downward direction whereas the torque due to the weight of the person is also acting in the downward direction whereas the tension the torque due to the tension in the string the t sine theta component of the tension in the string is acting in the direction opposite to the torque due to this two that is the weight of the beam and the weight of the person so therefore we can from here find the value of t that is the tension in the given cable tension in the given cable equal 300 multiplied 2 .5 plus 600 multiplied 1 .5 divided by 5 multiplied sine 53.
03:58
So the answer to this is about 413 newton.
04:03
So the tension in the given cable, this is the tension in the cable.
04:06
So this t here is the tension in the cable.
04:11
Now we have found the tension in the cable so that we can find the value of the force of reaction of the wall on the beam that is our objective.
04:21
So this gives the value of the tension in the cable, this value over here...