A 200 kg floodlight in a park is supported at the end of a...

A 20.0-kg floodlight in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole as shown in Figure P12.18. A cable at an angle of $\theta=30.0^{\circ}$ with the beam helps support the light. (a) Draw a force diagram for the beam. By computing torques about an axis at the hinge at the left-hand end of the beam, find (b) the tension in the cable, (c) the horizontal component of the force exerted by the pole on the beam, and (d) the vertical component of this force. Now solve the same problem from the force diagram from part (a) by computing torques around the junction between the cable and the beam at the right-hand end of the beam. Find (e) the vertical component of the force exerted by the pole on the beam, (f) the tension in the cable, and (g) the horizontal component of the force exerted by the pole on the beam. (h) Compare the solution to parts (b) through (d) with the solution to parts (e) through (g). Is either solution more accurate?

Transcript

00:01 In this problem, we are told we have a floodlight here that's attached to a canilever beam that's attached to a pole that's supporting it above the ground.

00:13 Then we also have a cable here that helps to support this canoliver.

00:19 Unfortunately, well, this really, i guess i shouldn't really call this a canilever.

00:23 They've drawn it as a canoliver, but that's not what they've assumed in the problem.

00:27 What we know is that this angle here is, 30 degrees and this guy weighs 20 kilograms.

00:35 So if we draw a force by a diagram for this beam here we have the tension from the cable, the weight of the floodlight, reaction forces from where it's attached to the pole and as they've drawn it there's a flange here which actually would create a moment at this point so that if this cable were broken this this beam would still possibly support itself because of this flange here.

01:05 But in the problem what they've assumed is that this is a pivot point so that if we cut this then this would obviously just rotate down.

01:15 So if we some moments about this point we get minus m so i've kept m in the problem for the time being so counterclockwise is positive so we have minus m and then t um t sign theta is the vertical component of the tension in the cable and so we multiply that by l to get the moment from that and then we have m minus mgl from the weight of the floodlight.

01:45 And we know everything in here well except for m and t and so we get t equals m over l plus mg all times cosat echin of theta and we'd be stuck there unless we assume that m is and if we do we get that t equals m g co -ssequent of theta which is 392 newton now to get the components and also to find these forces here the the x component of t is t sine of theta and again i guess i've taken this way to be positive but we could always again make this a negative and change the direction that we've defined to be positive.

02:33 But anyway, that's what we find for that and that means that f of x equals, so let's actually do that.

02:44 Let's say that this is positive is in the normal direction.

02:48 So this would be minus f of x.

02:50 Because if we draw some of the forces in the x direction, we just have this and this.

02:57 So the x component of t has to be offset by f of x.

03:01 So f of x becomes 196 newtons...