Question

The figure shows a uniform, horizontal beam (length = 10 m, mass = 60 kg) that is pivoted at the wall, with its far end supported by a cable that makes an angle of 51 degrees with the horizontal. If a person (mass = 20 kg) stands 3.0 m from the pivot, what is the tension in the cable?

Answer 1

First - resolve the tension in the rope into horizontal and vertical components.

Then take moments about the pivot point (i.e. the left end of line)

The mass act down , and th evertical component of the tension acts up.

Each has its own 'moment distance'

These 2 moments must be equal (and opposite) so you can calculate the vertical component of the tension.

Since you have the angle you can then calculate the tension.

Answer 2

@PsiSquared

Answer 3

Hi, @EducateMe. The first thing to note is that the system has to be in static equilibrium, which means that the moment about the base of the beam has to be zero (as @MrNood mentioned). In equation form that would be:\[M _{base}=M _{man}+M _{cable}\]Mbase is the moment around the base; Mman is the moment around the base generated by the man; and Mcable is the moment around the base generated by the cable. In this case, the moments are moments of forces, which means they are torques, so our equation can be rewritten as:\[T _{sys}=T _{man}+T _{cable}=0\]Now, remember that torque is just:\[T=Fd\]where T is torque; F is force; and d is distance. Also remember that torque can be positive or negative, with positive torque causing a counterclockwise rotation; and negative torque causing clockwise rotation.

Now consider the drawing I've attached. You see the man, his distance and the force he generates, and you also see the vertical force generated by the cable tension (i.e. the vertical component of the cable tension). All that's left undone is to figure out what that vertical component is. I've drawn the the cable tension and its component forces in a triangle. Knowing that angle, 51° and what the vertical component has to be, you should be able to find the tension.

Does this help?

Answer 4

Note that it's really not necessary to even consider positive and negative torque. If you solve the torque equation I gave you, the equation will automatically give you the right sign. The sign will indicate that Tcable acts in the opposite direction of Tman.

Answer 5

Ok, where did I wrong here?


\[M=5*60+Tsin(51)10=0\]

\[tsin(51)10=-300\]

\[T=\frac{ -300 }{ \sin(51)10 }\]

Answer 6

You haven't calculated a force for the man, and a force is what you need. You have md instead of mgd. T is a force.

Answer 7

Wait. I've left something out of the drawing. The total moment about the base of the beam is going to be:\[M _{system}=M _{man}+M _{beam}+M _{cable}\]And if we expand that equation we get:\[M _{system}=0=m _{man}d _{man}g+m _{beam}d _{cm-beam}g+T _{y-cable}d _{cable}\]where dcm is the distance to the center of mass of the beam. We can't forget that the beam is exerting a torque about its base (Picture only the beam being there with the one end fixed. The beam would rotate about that fixed point, wouldn't it?).Now adding what you got for Ty_cable we get:\[0=m _{m}d _{m}g+m _{b}d _{b-cm}g+Tsin(51°)d _{c}\]From now on I won't play with dogs while answering questions.

Answer 8

Is this helpful?

Answer 9

\[0=60*5+20*3+Tsin(51)10\]

What does the cmg mean? Am I suppose to subtract 10-3?
I placed a 5 instead of a 10 because it is the center mass, is that correct?
Do I need to do the same with the 3?
I'm guessing the cable (dc) would be 10

Answer 10

First it appears you've tried to divide g out of the equation. If you do that then you get the following:\[0=m _{m}d _{m}+m _{b}d _{b-cm}+\frac{ Tsin(51°)d _{c} }{g }\]db-cm is the distance from the pivot point at the end of the beam to the beam's center of mass (thus "cm"). g is the acceleration of gravity. You are correct, the distance from the pivot point to where the cable tension acts, dc, is 10m.

Answer 11

To be clear mm is the mass of the man; dm is the distance from the pivot point of the beam to the man; mb is the mass of the beam; db-cm is the distance to the beam's center of mass; dc is the distance to where the cable's tension acts on the system; and g is the acceleration of gravity.

Answer 12

mm = 20 kg
dm =3m
mb=60kg
db-cm=5m (10m originally but changed to 5 b/c of center mass
dc=10m
g=9.8m/s^2

0=mm*dm+mb*db-cm*g+Tsin(51)dc

\[20*3*9.8+60*5*9.8+Tsin(51)10=0\]

Is this right?

Answer 13

That looks right.

Answer 14

I'm getting -453.969 even thought my options are

a. 0.08 kN
b. 0.15 kN
c. 0.5 kN
d. 0.36 kN
e. 0.53 kN

Am I supposed to move the decimal to the left twice?

Answer 15

Your answer is in Newtons. The possible answers are all in kilo Newtons. 1kN=1000N.