# AP Physics B Torque

1. Oct 9, 2007

### koolkris623

A uniform steel beam of length 5.50 m has a weight of 4.50 103 N. One end of the beam is bolted to a vertical wall. The beam is held in a horizontal position by a cable attached between the other end of the beam and a point on the wall. The cable makes an angle of 25.0° above the horizontal. A load whose weight is 12.0 103 N is hung from the beam at a point that is 4.60 m from the wall.

(a) Find the magnitude of the tension in the supporting cable.
N
(b) Find the magnitude of the force exerted on the end of the beam by the bolt that attaches the beam to the wall.
N

my attempt:
ETorque = 0
Fl(D) + Wb (.5 D) - Ty(D) = 0
12.0E3N(4.60 m) + 4.50E3N(2.75) - Ty(6.069m) = 0
Ty = 11134.457 N

Find Tx : tan25 = 11134.457 / x
Tx = 23877.913 N

T = sqrt(23877.913^2 + 11134.457^2)
T = 26346.36 N

Efx = 0
23877.913 N - Fx = 0
Fx = 23877.913 N

EFy = 0
Fy + Ty-Fl _fb = 0
11134.457 N - 12E3 N - 4.5 E3 N + Fy = 0
Fy = 5365.543 N

F = sqrt(23877.913^2 + 5365.543^2)
F = 24473.327 N

When i tried the T and F in the online submission, there were wrong both times, and now im on my last try. Can someone please help me??

2. Oct 9, 2007

### Staff: Mentor

Where did the distance 6.069 m come from?

3. Oct 9, 2007

### koolkris623

that is the length of the cable..i did 5.5 / cos25

4. Oct 9, 2007

### Staff: Mentor

When calculating torques use the same pivot point for all forces. You want the length of the beam, not the length of the cable.

5. Oct 9, 2007

### koolkris623

they gave the lenght of the beam..5.5 m

6. Oct 9, 2007

So use it!

7. Oct 9, 2007

### koolkris623

ETorque = 0
Fl(D) + Wb (.5 D) - Ty(D) = 0
12.0E3N(4.60 m) + 4.50E3N(2.75) - Ty(6.069m) = 0
i used it here.....................^ the T is for the cable thats added to the beam

8. Oct 9, 2007

### Staff: Mentor

That's true. You need to use distance from the pivot point for every force acting on the beam: its weight (in the middle), the load (at 4.60 m), the the tension in the cable (which is applied to the end of the beam).

9. Oct 9, 2007

### koolkris623

ya but then didnt i do the problem right...it says im not getting the correct answer...so im lost and i have only one try left

10. Oct 9, 2007

### Staff: Mentor

No, you didn't do the problem correctly--you used the wrong distance when you calculated the torque due to the tension. So of course you'll get the wrong answer until you use the correct distance and do it over.

11. Oct 9, 2007

### koolkris623

Doc Al...I cant do it again..cause i only have one more trial left cool ill try the site

12. Oct 9, 2007

### koolkris623

ok i got the correct answers using the website you gave me

13. Oct 9, 2007

### Staff: Mentor

I don't understand what you mean by "can't do it again". Just solve the problem on paper and post your solution here. Someone will check it.

(If you don't do it again, how will you ever get the right answer?)

14. Oct 9, 2007

### koolkris623

can you explain to me what the website did that i didnt..cause i need to know how to do this for a test

Correct Ans:
29072.0130875 N for tension

Sum of forces =
26682.98896

15. Oct 9, 2007

### Staff: Mentor

But I've already told you what you did wrong--twice!

16. Oct 9, 2007

### koolkris623

oh hahaha i c the mistake lol i wasnt reading properly thanks for EVERYTHING!!!!