Go on. Explore!

lvl 0 • 0 xp

lvl 1

Home Market Decks Interests

AP Physics 1 - Torque Review

memorize.ai (lvl 286)

(0)

21 cards

()

Section 1

Preview this deck

Torque

Front

1 / 21

Physics

0.0

0 reviews

5			0
4			0
3			0
2			0
1			0

Active users

All-time users

Favorites

Last updated

6 years ago

Date created

Mar 1, 2020

Cards (21)

Section 1

(21 cards)

Torque

Front

Measure of the ability of the force to cause rotation/ Measurement of a ROTATION caused by a FORCE

Back

torque units

Front

Nm(Newton-meters) vector

Back

The total length of the see-saw is 8m. Stefani is at the very end of the see-saw with a weight of 360 N. The second child (Stefan) has a mass of 790 N. The see-saw is balanced. What is the net torque?

Front

0 Nm

Back

Torque is affected by

Front

-the magnitude of force applied -length of moment arm from axis of rotation to point of force application -orientation of the line of action of the force (angle between the two)

Back

What equation do you use if distance and force aren't perpendicular

Front

Torque(Nm) = rF sin θ

Back

Fulcrum/ pivot point

Front

Point of ROTATION *turning point of a wrench

Back

equilibrium

Front

a steady state due to the equal action of opposing forces

Back

The total length of the see-saw is 8m. Stefani is at the very end of the see-saw with a weight of 360 N. The second child (Stefan) has a mass of 790 N. Where should Stefan sit to balance Stefani's torque?

Front

1.82 m from the center Solution 8/2 =4 each side of see saw is 4 m long (360)(4) = (790) (x) Solve for x

Back

Rotational Equilibrium

Front

The sum of all the torque forces is equal to 0.

Back

torque acting on an object tends to produce

Front

rotation

Back

what determines the amount of torque produced by a torque?

Front

the amount of force applied as force application increases so does torque And/or The distance the force is applied at.

Back

Center of mass formula

Front

Where Xcm = center of mass, M1 = mass 1 (kg) x1 = distance (m)

Back

Center of Mass

Front

The point at which the mass of an object appears to act.

Back

counter clockwise rotation

Front

+ torque

Back

How much torque is created by applying 20 N of force to a 1.6 m wide door?

Front

32 Nm Solution Torque = distance x force = 20 x 1.6

Back

Translational Equilibrium

Front

The sum of the forces is equal to 0.

Back

Torque Equation

Front

Torque (Nm) = force (N) * lever arm (m) or T=rFsin(theta)

Back

The total length of the see-saw is 8m. Stefani is at the very end of the see-saw with a weight of 360 N. The second child (Stefan) has a weight of 790 N. Their little brother, Julian, joins Stefani on her side. Julian has a weight of 150 N and is sitting 1.4 m from the center. Now where should Stefan sit in order to balance out Stefani and Julian?

Front

2.2.1 m from the center Solution 8/2 =4 each side of see saw is 4 m long (360) (4) + (150 )(1.4)= (790) (x) Solve for x

Back

Lever arm

Front

The PERPENDICULAR DISTANCE between the FULCRUM & the point of application of the FORCE

Back

What equation do you use if distance and force are perpendicular?

Front

Torque(Nm) = rF sin θ

Back

Clockwise rotation

Front

- torque

Back