Force Calculations

force push pull wrestlers

 

Force is push or pull.

 

Forces on an object are usually balanced (if unbalanced the object accelerates):

forces balanced
  forces unbalanced
Balanced   Unbalanced
No Acceleration   Acceleration

Example: The forces at the top of this bridge tower are in balance (it is not accelerating):

suspension bridge

The cables pull downwards equally to the left and right, and that is balanced by the tower's upwards push. (Does the tower push? Yes! Imagine you stand there instead of the tower.)

We can model the forces like this:

suspension bridge forces

And when we put them head-to-tail we see they close back on themselves, meaning the net effect is zero:

suspension bridge forces
The forces are in balance.

Forces in balance are said to be in equilibrium: there is also no change in motion.

Free Body Diagrams

The first step is to draw a Free Body Diagram (also called a Force Diagram)

Free Body Diagram: A sketch where a body is cut free from the world except for the forces acting on it.

In the bridge example the free body diagram for the top of the tower is:

suspension bridge free body
Free Body Diagram

It helps us to think clearly about the forces acting on the body.

Example: Car on a Highway

What are the forces on a car cruising down the highway?

car moving

The engine is working hard, so why doesn't the car continue to accelerate?

Because the driving force is balanced by:

Like this:

car forces: driving, weight, air, tyres
Free Body Diagram

W is the car's weight,

R1 and R2 are the rolling resistance of the tires,

N1 and N2 are the reaction forces (balancing out the car's weight).

Note: steel wheels (like on trains) have less rolling resistance, but are way too slippery on the road!

Calculations

Force is a vector. A vector has magnitude (size) and direction:

vector magnitude and direction

We can model the forces by drawing arrows of the correct size and direction. Like this:

Example: Admiring the View

Brady stands on the edge of a balcony supported by a horizontal beam and a strut:

force man on beam with strut at 60 degrees

 

He weighs 80kg.

What are the forces?

 

Let's take the spot he is standing on and think about the forces just there:

free body diagram man on beam with strut at 60 degrees

His Weight

His 80 kg mass creates a downward force due to Gravity.

Force is mass times acceleration: F = ma

The acceleration due to gravity on Earth is 9.81 m/s2, so a = 9.81 m/s2

F = 80 kg × 9.81 m/s2

F = 785 N

The Other Forces

The forces are balanced, so they should close back on themselves like this:

force beam strut

We can use trigonometry to solve it.
Because it is a right-angled triangle, SOHCAHTOA will help.

For the Beam, we know the Adjacent, we want to know the Opposite, and "TOA" tells us to use Tangent:

tan(60°) = Beam/785 N

Beam/785 N = tan(60°)

Beam = tan(60°) × 785 N

Beam = 1.732... × 785 N = 1360 N

For the Strut, we know the Adjacent, we want to know the Hypotenuse, and "CAH" tells us to use Cosine:

cos(60°) = 785 N / Strut

Strut × cos(60°) = 785 N

Strut = 785 N / cos(60°)

Strut = 785 N / 0.5 = 1570 N

Solved:

force beam 1360 strut 1570

Interesting how much force is on the beam and strut compared to the weight being supported!

Torque (or Moment)

What if the beam is just stuck into the wall (called a cantilever)?

force man cantilever

There is no supporting strut, so what happens to the forces?

The Free Body Diagram looks like this:

force cantilever free body diagram

The upwards force R balances the downwards Weight.

With only those two forces the beam will spin like a propeller! But there is also a "turning effect" M called Moment (or Torque) that balances it out:

Moment: Force times the Distance at right angles.

We know the Weight is 785 N, and we also need to know the distance at right angles, which in this case is 3.2 m.

M = 785 N x 3.2 m = 2512 Nm

And that moment is what stops the beam from rotating.

moment on fishing rod

 

You can feel moment when holding onto a fishing rod.

As well as holding up its weight you have to stop it from rotating downwards.

Friction

Box on a Ramp

forces box on 20 degress incline: W, f, R

The box weighs 100 kg.

The friction force is enough to keep it where it is.

The reaction force R is at right angles to the ramp.

The box is not accelerating, so the forces are in balance:

force diagram:  W, f, R

The 100 kg mass creates a downward force due to Gravity:

W = 100 kg × 9.81 m/s2 = 981 N

 

We can use SOHCAHTOA to solve the triangle.

Friction f:

sin(20°) = f/981 N

f = sin(20°) × 981 N = 336 N

Reaction N:

cos(20°) = R/981 N

R = cos(20°) × 981 N = 922 N

And we get:

force diagram:  W=981N, f=336N, R=922N

Tips for Drawing Free Body Diagrams

Sam and Alex Pull a Box

The calculations can sometimes be easier when we turn magnitude and direction into x and y:

vector polar <=> vector cartesian
Vector a in Polar
Coordinates
  Vector a in Cartesian
Coordinates

You can read how to convert them at Polar and Cartesian Coordinates, but here is a quick summary:

From Polar Coordinates (r,θ)
to Cartesian Coordinates (x,y)
  From Cartesian Coordinates (x,y)
to Polar Coordinates (r,θ)
  • x = r × cos( θ )
  • y = r × sin( θ )
 
  • r = √ ( x2 + y2 )
  • θ = tan-1 ( y / x )

Let's use them!

vector example: 2 people pulling box

Example: Pulling a Box

Sam and Alex are pulling a box (viewed from above):

What is the combined force, and its direction?

Let us add the two vectors head to tail:

vector example: 200 at 60, 120 at 45

First convert from polar to Cartesian (to 2 decimals):

Sam's Vector:

Alex's Vector:

Now we have:

vector example: force components

Add them:

(100, 173.21) + (84.85, −84.85) = (184.85, 88.36)

That answer is valid, but let's convert back to polar as the question was in polar:

And we have this (rounded) result:
vector example: forces in trianle

And it looks like this for Sam and Alex:
vector example: combined forces

They might get a better result if they were shoulder-to-shoulder!