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# Torque - PowerPoint PPT Presentation

Torque. It is easier to open a door when a force is applied at the knob as opposed to a position closer to the hinges. The farther away the force, the more torque there is. Torque – the force(s) that cause an object to rotate

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## PowerPoint Slideshow about ' Torque' - yvette-burns

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Presentation Transcript

It is easier to open a door when a force is applied at the knob as opposed to a position closer to the hinges.

The farther away the force, the more torque there is.

Torque is the product of the force and the perpendicular distance from the axis of location

The force is extended to meet the moment arm which is drawn perpendicular to force through the hinge.

τ = rsinθF

Diagram B

The force is resolved into components and the perpendicular component is used with the original moment arm.

τ = rFsinθ

Either method for calculating torque is acceptable

τ = rFsinӨ

τ = torque (mN)

r =moment arm (m)

F = force (N)

The force and the moment arm should always be perpendicular to each other.

Two forces (Fa = 30 N and Fb = 20 N) are applied to a meter stick which rotates around its left end. Which force exerts the greater torque? What would be the net torque if they were applied together?

Calculate the net torque acting on two wheels that are attached. Ra is 30 cm and rb is 50 cm. Each force applied is 50 N

Rotational Statics attached. R

“Static” problems are ones where the objects in question are stationary.

Is there a net force acting on the traffic light?

No…the light is stationary.

ΣFx = 0 and ΣFy = 0

If the two forces are equal… attached. R

Is there a net force?

No

Will it move?

It rotates because of the torge

Στ ≠ 0

Two children are playing on a seesaw which has a mass of 2.0 kg.. If child A (m = 30 kg) sits 2.5 m away from the center, at what distance must child B (m = 25 kg) sit in order to balance the seesaw?

• When doing rotational static problems…. kg.. If child A (m = 30 kg) sits 2.5 m away from the center, at what distance must child B (m = 25 kg) sit in order to balance the seesaw?

• Draw a FBD showing all forces.

• Choose a convenient coordinate system

• Write the net force equations for the forces (x and y). They should all be equal to zero.

• Write the net torque equation. It should also be equal to zero. Choosing a strategic axis of rotation eliminates unknown forces.

• Solve for the unknowns.

Two children are playing on a seesaw which has a mass of 2.0 kg.. If child A (m = 30 kg) sits 2.5 m away from the center…

At what distance must child B (m = 25 kg) sit in order to balance the seesaw?

What is the normal force acting on the seesaw by the fulcrum?

A uniform beam, 2.20-m long with a mass of 25.0 kg (m), is mounted by a hinge on a wall. The beam is held horizontally by a cable that makes an angle of 30 degrees as shown. The beam supports a sign with a mass of 28.0 kg (M). Determine the force exerted by the hinge on the beam and the tension force in the cable.

A 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press 5.0 m from the right support column. Calculate the force on each support column.

This object is moving in a circular path at a constant speed. What is causing it to move in a circular path?

The string.

The string is exerting a force on the object

Net force? Acceleration?

Yes, even though the speed is constant.

The direction is changing; therefore, the velocity is changing.

It is accelerating

Planets move around the Sun in a circular path because of speed. What is causing it to move in a circular path?gravity.

The planets have a radial or tangential speed around the orbit.

The force that is causing the motion is directed toward the center.

a speed. What is causing it to move in a circular path?R = v2/r

aR = radial (centripetal) acceleration (m/s2)

r = radius of circle (m)

Centripetal forces cause centripetal accelerations. Both are vector quantities pointed toward the center of the circle.

A centripetal force is not a new type of force. Many of the forces we have discussed are centripetal forces when they cause objects to move in circles.

Σ are vector quantities pointed toward the center of the circle. FR = maR = mv2/r

Σ FR = net centripetal force (N)

m = mass (kg)

aR = centripetal acceleration (m/s2)

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second. What is the centripetal acceleration?

A 0.150-kg ball on the end of a 1.10-m cord is swung in a vertical circle. What is the minimum speed the ball must have a the top of its arc so that the ball continues moving in a circle?

What would be the tension of the cord at the bottom if the ball is traveling twice the speed of the 1st part?

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 14 m/s. Assuming the pavement is dry (μs= 0.60), will the car follow the curve or will it skid?

What if the pavement was covered with ice (μs = 0.25)?

Banked Turns at a speed of 14 m/s. Assuming the pavement is dry (

Banked turns assist in helping cars navigate turns when there is not sufficient friction for excessive speeds. How much of an angle is needed so that friction is unnecessary?

How much of a banked turn would be needed for a 1000-kg car to steer through a 50-m radius turn at a speed of 14 m/s?

The mass is irrelevant. Why?

Newton’s Law of Gravitation at a speed of 14 m/s. Assuming the pavement is dry (

Newton wondered why if an apple falls out of a tree toward the Earth why the moon doesn’t fall toward the Earth also.

.

The moon actually falls, but because it is not stationary (it has an initial radial speed), it follows a curved path around the Earth.

Newton theorized that the Moon did not get attracted with the same force as the apple nor did it fall with the same gravitational acceleration. Why not?

1. The moon was much farther away from the Earth than an apple on the surface.

2. The moon was much larger than the apple.

F (it has an initial radial speed), it follows a curved path around the Earth.g = Gm1m2/r2

G = gravitational constant = 6.67 x 10-11 Nm2/kg2

m1 = mass #1 (kg)

m2 = mass #2 (kg)

r = distance between centers of mass (m)

Newton’s Law of Universal Gravitation applies between any two objects that have mass regardless of size.

A 50-kg woman sits 0.5 meters away from a 75-kg man. What is the gravitational force between the two people?

To find out the acceleration of something falling on the surface of the Earth, Newton’s 2nd Law can be applied.

Fg = Gm1m2/r2

mg = GmEm/rE2

g = GmE/rE2

g = gravitational acceleration (m/s2)

mE = mass of Earth (kg)

r = radius of Earth (m)

How is a ball’s gravitational surface of the Earth, Newton’s 2acceleration on Mt. Everest (altitude = 8850 m) different from its gravitational acceleration at sea level?

Satellite Motion surface of the Earth, Newton’s 2

Satellites are objects that orbit the Earth. They are given a radial speed that counteracts the effects of gravity so that they maintain either a circular path.

People that are in satellite orbit experience “apparent weightlessness”.

Different from real weightlessness because the satellite and occupants are actually falling, but the radial speed keeps them in orbit.

It is like the apparent weightlessness of an elevator moving down.

Astronauts train in a similar weightless environment in a “zero g” airplane.

Plane makes large parabolic turns with 30-s intervals of weighlessness.

What do they experience at the bottom of the curve?

Calculate the speed of a satellite moving in stable circular orbit about the Earth at a height of 3600 km. Assume the radius of the Earth to be 6.38 x 106 m and the mass to be 5.98 x 1024 kg.
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