# Rotational Dynamics of a cable car

1. Jul 24, 2008

### GeoKenshin

Hi. I've been having trouble with 2 problems and I've worked them both 2 or 3 times. Here's my problems and my work. Any help will be greatly appreciated.

1.
1. The problem statement, all variables and given/known data

In San Francisco a very simple technique is used to turn around a cable car when it reaches the end of its route. The car rolls onto a turntable, which can rotate about a vertical axis through its center. Then, two people push perpendicularly on the car, one at each end, as in the drawing. The turntable is rotated one-half of a revolution to turn the car around. If the length of the car is 9.80 m and each person pushes with a 185 N force, what is the magnitude of the net torque applied to the car?

2. Relevant equations

T=F (LcosTheta)

3. The attempt at a solution

I also used the (F1 + F2) L equation but i just got the same answer with a positive sign

T=F (LcosTheta)

T = 370(9.8cos180) = -3626

2.

1. The problem statement, all variables and given/known data

The drawing shows an outstretched arm (0.61 m in length) that is parallel to the floor. The arm is pulling downward against the ring attached to the pulley system, in order to hold the 126 N weight stationary. To pull the arm downward, the latissimus dorsi muscle applies the force vector M in the drawing, at a point that is 0.069 m from the shoulder joint and oriented at an angle of 29°. The arm has a weight of 44 N and a center of gravity (cg) that is located d = 0.26 m from the shoulder joint. Find the magnitude of vector M .

2. Relevant equations

T = -Wa(La) - Wd(Ld) + M (Lm) = 0

3. The attempt at a solution

I used the above equation to get M = [Na(La/Center of Gravity) + Nb(Lb)]/Lm

[44N(.26m) + 126N(.61m)]/[.69mcos29] = 146.316 N

That Answer isn't right and I'm really stuff on what I could be doing wrong.

Thanks in advance for any assistance.

#### Attached Files:

File size:
12.3 KB
Views:
44
• ###### arm.gif
File size:
35.6 KB
Views:
49
2. Jul 24, 2008

### cryptoguy

For #1, T=F (LcosTheta) is true but what is theta in this case? It's NOT 180. The force is being applied at a 90 degree angle to the cable car.
Additionally, you need to consider each force separately, the distance would not be 9.8 m. You need the distance from each force to the center.

For #2, you should be considering the vertical component of M, so I'd reconsider the cos :) oh and this is porlly a typo but you have .69 written instead of .069

3. Jul 24, 2008

### GeoKenshin

if the angle is 90 degrees for #1 then Cos of 90 degrees is 0 and that would make the net torque zero and that can't be right can it?

for #2 I got the answer 2639.61 N after using sine instead of cosine but i don't think that is correct either i got that yesterday as well and it wasn't correct. So would i use tan then for some reason?

4. Jul 24, 2008

### Staff: Mentor

No it cannot. The correct formula for torque is $\tau = FL\sin\theta$. Note that L is the distance to the axis.

5. Jul 24, 2008

### GeoKenshin

Okay i have the first problem correct now so thank you. Is there something wrong with my formulas for the second problem? I can't seem to get the right answer.

T = -Wa(La) - Wd(Ld) + M (Lm) = 0

M = [Na(La/Center of Gravity) + Nb(Lb)]/Lm

= [ 44(.26) + 126(.61) ] / (.069sin29) = 2639.61376 N

6. Jul 24, 2008

### Staff: Mentor

Your signs are off. The torque due to force M and the weight of the arm both act clockwise and thus must have the same sign.

7. Jul 24, 2008

### GeoKenshin

got it! thanks.