Torque Rotational Eq

[laczlacylaci]

Physics has been so long, and I don't remember anything except T=rFsin(theta)
In the answer key, it says the 10kg weight is applying a torque of approximately 35Nm. <-how did they calculate this: (35cm to 35Nm?)

F=T/(rsin(theta)) eq manipulation to solve for F
T=35Nm; r=.02m; theta=75`

---

Members don't see this ad.

 

[theonlytycrane]

Ignore the angle and think of the 20mm mark (where the arrow saying 75 degrees is pointing) as the fulcrum.

To balance at the fulcrum, the torque of the weight (down) should equal the torque of the elbow (also down).

Use T = r * F since both torques are straight down and sin(90) = 1.

F(elbow) * 20 mM = 10 kg * 10 m/s^2 * (35 cm - 20 mm ~ which is basically 35 cm).

They get 35 N*m from 35 cm * 10 kg * 10 m/s^2 above ^.

 

[majikarp]

The torque created by gravitational force of the weight = rfsintheta = (35cm)(10kg x g)(sin 90) = 35N
The torque that must be created by the bicep going counter clockwise = r f sin theta = (20mm)(X)(sin 75) = .02X
.02X = 35
X = 1750N (a little bit off since I approximated sin 75 = 1 when its actually .96 or something)