Chapter+8+Rotational+Equilibrium

=Chapter 8 Rotational Equilibrium=
 * By:Emma Weikel**

-measured in Newton meters (Nm)
 * Torque**-force that causes rotation about an axis

The torque arm is perpendicular distance from force to the pivot point. -The longer the torque arm, the easier it is.
 * Torque**=(Force)(length of torque arm)


 * EXAMPLE Problem:**
 * Using a wrench to unscrew a bolt with a torque arm of 25 cm and applying a force of 10 N downward. Determine the torque applied. (Be sure to convert 25 cm to m)**


 * torque=(10 N)(0.25 m)** **=2.5 Nm**

If not perpendicular, use an angle to find torque. --> **Use the same formula above, just add sin(theta)**


 * EXAMPLE Problem:**
 * A wrench of 0.5 m long is applied to a nut with a force of 80 N. Because of the cramped space, the force must be exerted upward at an angle of 60 degrees with respect to a line from the bolt through the end of the wrench. How much torque is applied to the nut?**


 * torque=(80 N)(0.5)(sin 60)** **=34.64 Nm**

This is essential when finding **equilibrium** of an object.
 * Center of Mass**-an objects weight varies; it is where the weight is concentrated.

- the sum of F=0
 * 2 Types**
 * Translation**- straight line
 * Rotational**- sun of torque=0

This can help understand torque!... []

__**Multiple Choice Questions**__

 * 1. The torque arm for nut and wrench is...**
 * a. The whole thing**
 * b. From the handle to the bolt**
 * c. From the pivot point to handle**
 * d. Just the handle**


 * 2. Another unit that is equivalent to a Newton meter (Nm) is...**
 * a. ftlb**
 * b. kglb**
 * c. Nft**
 * d. None of the above**


 * 3.For rotational equilibrium, the sum of torque that is moved clockwise is equal to...**
 * a. the same at rest with increased angular velocity**
 * b. the same at rest or constant angular velocity**
 * c. different at rest with decreased angular velocity**
 * d. different at rest with constant angular velocity**


 * 4. A 3 kg mass is placed 2 m to the right of the pivot point of a see-saw. What is the magnitude of the torque applied?**
 * a. 6 Nm**
 * b. 58.8 Nm**
 * c. 20.5 Nm**
 * d. 5.7 Nm**


 * 5. The pivot point is at the hinges of the door opposite to where you were pushing it. The force you used was 55 N at a distance of 3.5 m from the pivot point. You hit the door perpendicular to its plane so the angle between the door and direction of force was 93 degrees. Find the torque in this example.**
 * a. 90 Nm**
 * b. 130 Nm**
 * c. 190.2 Nm**
 * d. 192.2 Nm**