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TOUGH Static Equilibrium Question

  • Thread starter Hotsuma
  • Start date
1. Homework Statement

A traffic light hangs from a pole as shown in the figure (Intro 1 figure, see link: http://session.masteringphysics.com/problemAsset/1058163/3/GIANCOLI.ch12.p21.jpg" [Broken] ) . The uniform aluminum pole AB is 7.20 m long and has a mass of 11.0 kg. The mass of the traffic light is 20.5 kg.

Part 1) Determine the tension in the horizontal massless cable CD.

Part 2) Determine the vertical component of the force exerted by the pivot A on the aluminum pole.

Part 3) Determine the horizontal component of the force exerted by the pivot A on the aluminum pole.

2. Homework Equations

[tex]\sum{}F=0[/tex]

[tex]\sum\tau=0[/tex]

One of those should be F, for some reason TeX is not working for me...

3. The Attempt at a Solution

Okay, so first I calculate all the distances between the points:

CD = 5.72 m
AD = 6.868 m
DB = .332 m

The two theta values are:

37*
53*

From there I know the x components and y components of Force should be zero, and I find that the vertical component of force exerted by the pivot A is 309 N. This answer for Part 2 is correct. From then I tried to calculate the horizontal component by dividing that number by tangent of 37 degrees (or by multiplying by 53 degrees) but the answer, 410 N, is not correct. I know from the summation of forces on the x co-ordinate plane that the tension force should equal the horizontal component of force from the arm A, which is part 3. So part 1 = part 3. However, I don't know how to get any other information from this point, because summation of torques only can give me what I already know. Is this a problem with Mastering Physics and should I just double check my values?
 
Last edited by a moderator:

LowlyPion

Homework Helper
3,079
4
For 1) you really just need to consider the Σ T about the pivot.

You have 3 forces. The weight of the light at a moment arm of 7.2m and the Center of mass of the pole at 3.6m Added together then they are both acting clockwise about the pivot at cos 37.

The 3rd force is your Tension acting counter clockwise at a distance given by the problem as 3.8m. That makes the answer then just the Σ of the clockwise T / 3.8m.

You should be able to determine the remaining components then of the forces acting along the pole into the pivot, now that you know the 3 forces acting on it.
 

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