4. The figure shows a uniform, horizontal beam (length = 10...

4. The figure shows a uniform, horizontal beam (length = 10 m, mass = 35 kg) that is pivoted at the wall, with its far end supported by a cable that makes an angle of 51° with the horizontal. (a) If a person (mass = 80 kg) stands 3.0 m from the pivot, what is the tension in the cable? (b) What is the angle between the net force acting on the pivot and the beam?

Transcript

00:02 All right, so here we have a person on a beam that's supported by a cable that has tension in it.

00:10 And there is a hinge at a.

00:13 And our goal is to solve for the tension in the cable as well as the angle of the force at the hinge.

00:24 So let's go ahead and first solve for the tension.

00:27 We're going to do that by doing some of the torque about hinge a.

00:32 And we're going to go ahead and make clockwise positive.

00:36 So we have, oh, let me add, i've got some dimensions to add here.

00:41 So we've got everything on here.

00:43 Let's do it in red.

00:45 Okay, so what i'm going to do is make this distance x and the full length of the beam will be l.

00:54 And the values are given in the problem, and i'll go ahead and add them in as we solve the problem.

01:01 All right, so we are going to do some of the talk about hinge a.

01:08 So clockwise will count positive here.

01:10 So we've got the person, weight of the person times the distance from the axis of rotation at a.

01:19 We also have the weight of the beam.

01:23 And the way the beam is considered at its center of mass, so at l over 2.

01:28 And going counterclockwise for a four hour torque, we have ty, which is t, sine, theta, and that is applied a distance l -o -a, and that equals zero because we are not rotating.

01:48 Okay, so basically we want to solve for t.

01:51 So if i rearrange the equation for t, i'll get big mgx plus plus little m g l over 2, all over l sine theta.

02:04 That's just rearranging using algebra to get t.

02:07 We can now plug in our values, since we have values, we have 80 kilograms on the person on the beam, 9 .8 meters per second squared for g.

02:21 Our distance x is three meters.

02:25 Then we have the mass of the beam, which is 35 kilograms and g is 9 .8 meter per second squared running out of room so i've got to squish it in here and half of the length of the beam is five meters.

02:44 I'm going to divide that by the full length of the beam times sign of 51 degrees because that's our given angle for theta.

02:52 Whoops, that should be 51.

02:55 Okay, so when we plug that into our calculator, the tension is 523 but 3 to 5 newtons.

03:04 So that's part a, the tension in the cable.

03:08 All right, now let's do part b...

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