Transcript Torque
Torque
Torque is defined as the tendency to produce a change in rotational motion.
Torque is a twist or turn that tends to produce rotation. * * * Applications are found in many common tools around the home or industry where it is necessary to turn, tighten or loosen devices.
What makes something rotate in the first place?
TORQUE
How do I apply a force to make the rod rotate about the axis? Not just anywhere!
AXIS
TORQUE
• To make an object rotate, a force must be applied in the right place.
• the combination of force and the distance from the axis to the point of application (L) is called TORQUE. lever arm, L Axis Force, F
Torque = force times lever arm Torque = F
L
Torque example
L F What is the torque on a bolt applied with a wrench that has a lever arm of 30 cm with a force of 30 N?
Torque = F L = 30 N = 9 N m 0.30 m For the same force, you get more torque with a bigger wrench the job is easier!
Units for Torque
Torque is proportional to the magnitude of F and to the distance L from the axis. Thus, a tentative formula might be: t
= FL
Units: N m or lb ft t
=
(40 N)(0.60 m) = 24.0 N m, t
=
24.0 N m, 60 cm 40 N
Direction of Torque
Torque is a vector quantity that has direction as well as magnitude. Turning the handle of a screwdriver clockwise (negative) and then counterclockwise (positive) will advance the screw first inward and then outward.
Sign Convention for Torque
By convention, counterclockwise torques are positive and clockwise torques are negative.
Positive torque: Counter-clockwise, out of page cw ccw Negative torque: clockwise, into page
Line of Action of a Force
The
line of action
of a force is an imaginary line of indefinite length drawn along the direction of the force.
F 1 F 2 Line of action F 3
The Moment Arm
The
lever arm
distance from the line of action of a force to the axis of rotation.
of a force is the perpendicular F 1 F 2 L L L F 3
Calculating Torque
• Read problem and draw a rough figure.
• Extend line of action of the force.
• Draw and label Lever arm.
• Calculate the Lever arm if necessary.
• Apply definition of torque: t
= FL Torque = force x Lever arm
Torque
• If we know the angle between F and r, we can calculate torque!
t = F L – r is the total length – F is force L = r sin Hinge (rotates) t – = r sin F Direction of rotation is angle between F and r
L r
• The SI unit of torque is the Nm. F
Extend the line of action
Example 1: a 12-cm An 80-N force acts at the end of wrench as shown. Find the torque.
L • Extend line of action, draw, calculate L t
=
(80 N)(.12m sin 60 o ) = 8.31 N m
Net Torque
An object is in “Equilibrium” when: 1. There is no net force acting on the object
2. There is no net Torque
In other words, the object is
NOT experiencing
linear acceleration or
rotational acceleration.
a
v
t
t
0 0
What mass is needed for the levers to be in equilibrium?
Weights are attached to
8 meter
long levers
at rest
.
Determine the unknown
weights
below
??
20 N 20 N
??
??
20 N
What mass is needed for the levers to be in equilibrium?
Upward force from the fulcrum produces no torque (since r = 0)
r1 = 4 m r2 = 4 m 0 = -F 2 r 2 + F 1 r 1 0 = -(F 2 )(4)+ (20)(4)
F 2 = 20 N … same as F 1
F 1 = 20 N
F 2 =??
What mass is needed for the levers to be in equilibrium?
r1 = 4 m r2 = 2 m F 1 = 20 N
F 2 =??
(force at the fulcrum is not shown)
0 = -F 2 r 2 + F 1 r 1 0 = -(F 2 )(2) + (20)(4)
F 2 = 40 N
More interesting problems (the pivot is not at the center of mass) Masses are attached to an 8 meter long lever
at rest
.
The lever has a mass of 10 kg.
Determine the unknown
weight
below.
CM 20 N
??
More interesting problems (the pivot is not at the center of mass)
Trick: gravity applies a torque “equivalent to” (the weight of the lever)(R cm )
t cm =(mg)(r cm ) = (100 N)(2 m) = 200 Nm CM
??
20 N Weight of lever Masses are attached to an 8 meter long lever
at rest
.
The lever has a mass of 10 kg.
Masses are attached to an 8 meter long lever
at rest
.
The lever has a mass of 10 kg.
Determine the unknown
weight
below.
R1 = 6 m CM Rcm = 2 m R2 = 2 m 0 = -F 2 r 2 + F 1 r 1 + F cm R cm 0 = -(F 2 )(2) + (20)(6)+(100)(2) Fcm = 100 N
F 2 = 160 N
F1 = 20 N F2 = ??
Diving board A 8 meter long diving board with a mass of 40 kg.
a. Determine the downward force of the bolt.
bolt t cm = (392 N) 2 m = 784 Nm F 1 r 1 = F 2 r 2 + t cm
The Pivot point is not at the center of mass
Diving board A 8 meter long diving board with a mass of 40 kg.
a. Determine the downward force of the bolt.
(Balance Torques)
bolt R 1 = 2 F 1 R cm Nm = 2 r 1 = t cm = 784 F bolt = ? N F cm = 392 N F1 = 784Nm = 392 N 2m
Diving board A 4 meter long diving board with a mass of 40 kg.
b. Determine the upward force applied by the fulcrum.
(Balance Forces) F = 784 N
bolt The total upward force has to equal the total downward for the object to be stable. F bolt = 392 N F cm = 392 N
Remember:
An object is in “Equilibrium” when:
a. There is no net Torque
t 0
b. There is no net force acting on the object
F
0
Torque with two supports
Fr FL Fcm 1. Label all the forces 2. Choose a pivot point 3. Write the equation for net torques
Torque with two supports
F2 F1 Fcm Lcm L1 cm L4 Pivot point T net = 0 -F1 L1 + Fcm L cm + F2 (0)=0
Pick a pivot pt.
What is the force exerted at each end?
0 = F 1 F F 2 F 1 2 = 276 N + F 1 (0) -755 (1.40) – 167 (2.75) + F 2 = 755 + 167N = 646 N (5.50M) F 1 F 2