Calculating the amount of torque

(A confusing task)

A door

Hinge

Hinge

Knob

Door (Top View)

Hinge

knob

knob

How much torque?(Zero, a Little, or Large?)

How much torque?(Zero, a Little, or Large?)

How much torque?(Zero, a Little, or Large?)

How much torque?(Zero, a Little, or Large?)

How much torque?(Zero, a Little, or Large?)

How to calculate torqueMethod One

(“Adjust the force”)

F

F

Draw a vector from the axis to the point of application of force.

d

F

Extend the vector to make an angle between the force and d.

d

F

Complete the triangle to find the component of F that is perpendicular to d. This is the

only part of F that makes the torque.

F = Fsin d

F

= d·F = d·Fsin

F = Fsin d

F

How to calculate torqueMethod Two

{“Adjust the distance”} (more handy than you might think)

F

Draw the usual distance vector.

d

F

Extend the ForceVector to see the “Line of Force”

d

Fd

How far is the Line of Force from the pivot?

(We call this distance the ‘Lever Arm’ or “Moment Arm’)

d = dsin()

Fd

= F·d = F·dsin = d Fsin

dsin

Numerical Examples.For each, the force is 20 N, and the dimensions of the door are:

0.8 meters

0.3 m

Note: The diagonal distance is 0.854 meters.

Numerical Example A (Method One - Adjust Force)

F40˚

d = 0.8 m

Numerical Example A (Method One - Adjust Force)

50˚40˚

d = 0.8 mF = 20 N = 50˚

Numerical Example A (Method One - Adjust Force)

50˚40˚

d = 0.8 m

F = 20 N·sin(50˚) = 15.3N

F

Numerical Example A (Method One - Adjust Force)

50˚40˚

d = 0.8 mF = 20 N·sin(50˚) = 15.3N

F

= d·F= (0.8 m)·(15.3 N) = 12.3 Nm

Numerical Example A(Method Two - Adjust Distance)

F40˚

50˚40˚

90˚

d = 0.8 m

d = ?

Numerical Example A(Method Two - Adjust Distance)

F40˚

50˚40˚

d = (0.8 m)sin(˚) = 0.61 m

d = 0.8 m

Numerical Example A(Method Two - Adjust Distance)

F40˚

50˚40˚

d = (0.8 m)sin(˚) = 0.61 m

d = 0.8 m

= d·F= (0.61 m)·(20 N) = 12.3 Nm

Numerical Example BWill this be: More, Less, or the Same

amount of Torque?

d = 0.8 mF = 20 N

Numerical Example BMethod One - Adjust Force

90˚

Why is F = 20 N?

Numerical Example BMethod One - Adjust Force

90˚

= d·F= (0.8 m)·(20 N) = 16 NmIs this more, less, or the same?

Numerical Example B Method Two - Adjust Distance

d = 0.8 mF = 20 N

Numerical Example B Method Two - Adjust Distance

d = 0.8 mF = 20 N

Why is d = 0.8 m?

Numerical Example B Method Two - Adjust Distance

d = 0.8 mF = 20 N

= d·F= (0.8 m)·(20 N) = 16 Nm

Numerical Example C(shows why Method 2 is useful)

Numerical Example C(Method One - Adjust Force)

WARNING: This is a bad method for this problem.

20 N

0.8 m

0.3 m

Just sit back and see how much work this method is for this problem, and then how easy the othermethod is for the same problem.

Numerical Example CIn Method One, we find F.Draw in the component of

F that is to d.

d

D = √(0.82 + 0.32) = 0.854 m

Extend d. Find F.

d

Extend d. Find F.

d

F

= InvTan(Opp/Adj) = tan-1(0.8m/0.3m) = 69.4˚

Find F.

d

F

F = (20 N)sin(69.4˚) = 18.7 N

= d·F= (0.854m)·(18.7N)

= 16 Nm

d

F

This was a tough way to get the answer.

Numerical Example C(Method Two - Adjust Distance)

20 N

0.8 m

0.3 m

= F·dSo, find the distance from

the ‘Line of force’ to the pivot.20 N

0.8 m

0.3 m

d

20 N

0.8 m

0.3 m

= F·d

= (20 N)·(0.8 m) = 16 Nm

20 N

0.8 m

0.3 m

That was much better.

Addendum (skippable)A closer look at Method 1 in Example C

• In the next slide notice that the distance (d) comes from the width (W) of the door and the thickness (T) of the door:

€

d = W 2 + T 2

d

W

T

d = square root of …

In the previous slide, notice that

• The sine of the angle is Opp/Hyp• So

€

sinθ =W

W 2 + T 2

Calculate the torqueand see the simplification to the

old result• = d•F•sin

€

=FW€

= W 2 + T 2 ⋅F ⋅W

W 2 + T 2