TorqueTorquePhysicsPhysics

Montwood High SchoolMontwood High School

R. CasaoR. Casao

Torque As we have seen over and over again, a force As we have seen over and over again, a force

is a push or a pull, and an unbalanced force is a push or a pull, and an unbalanced force causes an object to move (to accelerate). causes an object to move (to accelerate).

All the motion we have dealt with so far has All the motion we have dealt with so far has moved objects from one point in space to moved objects from one point in space to another. This type of motion involving a another. This type of motion involving a displacement is called displacement is called translational motion.translational motion.

We will now consider another type of motion, We will now consider another type of motion, rotational motionrotational motion, where an object moves , where an object moves not from one place to another but rather not from one place to another but rather around some axis of rotation. around some axis of rotation.

Translation (linear) Motion

Translation + Rotation

Torque Forces are also necessary for Forces are also necessary for

rotational motion.rotational motion. The rotational effect of a force The rotational effect of a force

depends upon its point of depends upon its point of application.application.

TorqueTorque: is the rotational analog of : is the rotational analog of a force and measures the tendency a force and measures the tendency of a force to rotate a body around a of a force to rotate a body around a pivot (fulcrum). Actual rotation pivot (fulcrum). Actual rotation does not need to occur.does not need to occur.

Torque A torque is not a force.A torque is not a force. Pivot point: the point from which the Pivot point: the point from which the

lengths of all torque arms, also called lengths of all torque arms, also called moment arms, are measured.moment arms, are measured.

Once a pivot point is chosen, a Once a pivot point is chosen, a perpendicular line is drawn to each of perpendicular line is drawn to each of the lines along which the force vectors the lines along which the force vectors act on the object. Torque arm or act on the object. Torque arm or moment arm is the perpendicular moment arm is the perpendicular distance from the pivot point to the distance from the pivot point to the line of action of the force.line of action of the force.

Torque The line of action of the force passes The line of action of the force passes

through the point at which the force is through the point at which the force is applied and is parallel to the force.applied and is parallel to the force.

Calculating a torque always requires Calculating a torque always requires that you choose one point as your only that you choose one point as your only pivot and calculate all torques relative pivot and calculate all torques relative to it.to it.

Torque, the rotational effect of a force is Torque, the rotational effect of a force is the product of a force and the length of the product of a force and the length of its torque arm.its torque arm.

Torque Torque = Force applied Torque = Force applied ·· torque torque

armarm Mathematically: Mathematically:

Unit: m·N or N·mUnit: m·N or N·m When forces are not applied When forces are not applied

parallel (0parallel (000) to or perpendicularly ) to or perpendicularly (90(9000) to the torque arm, the ) to the torque arm, the mathematical expression for the mathematical expression for the torque is:torque is:

rFT

θsinrFT

Torque where where is the angle between the line of is the angle between the line of

action of the force and a line drawnaction of the force and a line drawn from the point of application of the from the point of application of the force to the pivot.force to the pivot.

The torque arm is once again the The torque arm is once again the perpendicular distance from the pivot perpendicular distance from the pivot to the line of action of the force.to the line of action of the force.

If the distance from the point of If the distance from the point of application of the force is r, then the application of the force is r, then the torque arm is one side of a right torque arm is one side of a right triangle in which r is the hypotenuse.triangle in which r is the hypotenuse.

Torque (simplest case)

F l

“Lever arm” or torque arm is perpendicular to force.

Units: N·m

Torque The torque for a force applied at the The torque for a force applied at the

pivot point is always zero. The moment pivot point is always zero. The moment arm for any force applied at the pivot arm for any force applied at the pivot point is zero.point is zero.

Direction of torque: Direction of torque: – To identify a torque as clockwise or To identify a torque as clockwise or

counterclockwise, imagine that the object is counterclockwise, imagine that the object is free to rotate around a stationary pivot point. free to rotate around a stationary pivot point.

– Imagine that the force producing the torque Imagine that the force producing the torque is the only force acting on the object. is the only force acting on the object.

– The direction in which the bar would rotate is The direction in which the bar would rotate is the direction of the torque. the direction of the torque.

Torque

Torque

How do we quantify torque when force is not perpendicular to the lever arm?

Define:• Line of Action:

The line along the force vector.

• Lever Arm:The distance between the line of action and the axis of rotation measured along a line that is perpendicular to both.

Torque

Zero torque if force isdirected toward center ofrotation.

Mechanics of Rigid Body

CMext amF

CMCM I

EQUATIONS OF MOTION OF A RIGID BODY

OO IIn the case of a rotation about a fixed axis and O is a point of this axis. In general, if O is a point of a inertial reference system

EQUILIBRIUM CONCEPT

Conditions for equilibrium

equilibrium = no acceleration, i.e. a particle remains at rest, or if in motion, moves with constant velocity.

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0

1. The net external force acting on the body must remain zero.

2. The net external torque about any point must remain zero. Any point can be considered because all of them are in a inertial reference system (non accelerated).

CONDITIONS FOR EQUILIBRIUM

Rotational EquilibriumRotational Equilibrium An object at equilibrium has no net An object at equilibrium has no net

influences to cause it to move, either influences to cause it to move, either in translation (linear motion) or in translation (linear motion) or rotation. The basic conditions for rotation. The basic conditions for equilibrium are: equilibrium are:

Rotational EquilibriumRotational Equilibrium A torque that tends to rotate A torque that tends to rotate

something counterclockwise is something counterclockwise is usually defined as positive; a usually defined as positive; a clockwise torque is defined as clockwise torque is defined as negative.negative.

Rotational equilibrium occurs when Rotational equilibrium occurs when the angular acceleration and the the angular acceleration and the sum ofsum of the torques is equal to zero.the torques is equal to zero. – Dynamic rotational equilibrium: Dynamic rotational equilibrium:

angular velocity is constant; angular angular velocity is constant; angular acceleration is zero.acceleration is zero.

Rotational EquilibriumRotational Equilibrium– Static rotational equilibrium: angular Static rotational equilibrium: angular

velocity and angular acceleration is velocity and angular acceleration is zero.zero.

Three factors determine how Three factors determine how effective a torque is in altering the effective a torque is in altering the rotational state of an object:rotational state of an object: – the magnitude of the force,the magnitude of the force,– the direction of the force, andthe direction of the force, and – the point of application of the force.the point of application of the force.

Problem-Solving TipsProblem-Solving Tips

All torques must be calculated All torques must be calculated about the same pivot.about the same pivot.

When a problem contains two When a problem contains two unknowns, mathematically remove unknowns, mathematically remove one of the unknowns by selecting one of the unknowns by selecting it as the pivot. This causes the it as the pivot. This causes the torque due to the applied force at torque due to the applied force at that location to be zero. that location to be zero.

Coupled ForcesCoupled Forces Couple: two forces of equal magnitude Couple: two forces of equal magnitude

acting in opposite directions in the same acting in opposite directions in the same plane, but not applied at the same point.plane, but not applied at the same point.

Although the forces are in opposite Although the forces are in opposite directions, each tends to rotate the body in directions, each tends to rotate the body in the same direction about the pivot.the same direction about the pivot.

The torque of a couple is the product of one The torque of a couple is the product of one of the forces and the perpendicular distance of the forces and the perpendicular distance between them.between them.

A couple can only be balanced by a second A couple can only be balanced by a second couple; the torques of the two couples must couple; the torques of the two couples must have equal magnitudes but opposite have equal magnitudes but opposite directions.directions.

The Lunch Tray Example

Tray: m = 0.2 kg; food: m = 1.0 kg; cup: m = 0.250 kg

What is F and T?

Your fingers and thumb both exert a force on the tray that acts to keep it in equilibrium.

The Lunch Tray: free-body diagram

Wt

Wf

Wc

F2

F1

Mechanics of Rigid Body: Statics

A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G.

Determine the components of the reactions at A and B.

• Determine B by solving the equation for the sum of the moments of all forces about A. Note there will be no contribution from the unknown reactions at A.

• Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components.

• Check the values obtained for the reactions by verifying that the sum of the moments about B of all forces is zero.

Free-body diagramFirst step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-body diagram.• Select the extent of the free-body

and detach it from the ground and all other bodies.

• Include the dimensions necessary to compute the torque arms of the forces.

• Indicate point of application and assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body.

• Indicate point of application, magnitude, and direction of external forces, including the rigid body weight.

A man raises a 10 kg joist, of length 4 m, by pulling on a rope.

Find the tension in the rope and the reaction at A.

SOLUTION:

• Create a free-body diagram of the joist. Note that the joist is a 3 force body acted upon by the rope, its weight, and the reaction at A.

• The three forces must be concurrent for static equilibrium. Therefore, the reaction R must pass through the intersection of the lines of action of the weight and rope forces. Determine the direction of the reaction force R.

Mechanics of Rigid Body.

Equilibrium Example: Ladder

WF=875 NWL=355 NdL=8.0 mdF=6.3 m

Find the forces that the wall and the ground exert on the ladder.

The Ladder

Free-body diagram:

Ladder: sum of torques is zero

The normal force exerted by the ground on the ladder and the frictional force exerted by the ground do not apply torques to the ladder.