Force Calculations

Two wrestlers pushing against each other

Force is push or pull.

Forces on an object are usually balanced

When forces are unbalanced the object accelerates:

Balanced Forces
Block with equal and opposite 10 Newton horizontal forces
No Acceleration
 
Unbalanced Forces
Block with unequal horizontal forces of 15 Newtons right and 10 Newtons left
Acceleration

Example: The forces at the top of this bridge tower are in balance (it isn't accelerating):

Suspension bridge with cables connected to a central tower

The cables pull downwards equally left and right, and that's 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:

Three force vectors at the top of a bridge tower: two pulling down-outward, one pushing straight up

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

Three force vectors connected head-to-tail to form a closed triangle
The forces are in balance.

Forces in balance are said to be in equilibrium: there's 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:

Free body diagram of the bridge tower top with three force arrows
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 driving on a road

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

Because the driving force is balanced by:

  • Air resistance (put simply: the air resists being pushed around),
  • Rolling resistance, also called rolling friction (the tires resist having their shape changed)

Like this:

Free body diagram of a car showing driving force, air resistance, weight, and normal forces
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!

Weight or Mass?

  • Mass is how much "stuff" (matter) is in an object. It is measured in kilograms (kg) and stays the same anywhere in the universe
  • Weight is the force of gravity pulling down on that mass. It is measured in Newtons (N) and changes depending on gravity (you weigh less on the Moon!)

Free Body Diagrams use forces, so we use the object's weight on the diagram.

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

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

Balcony supported by a horizontal beam and a strut at a sixty-degree angle

Jade weighs 80kg.

What are the forces?

Let's think about the forces at the end of the beam:

Free body diagram of balcony joint with downward, horizontal, and angled force vectors

Weight

Jade's 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:

Right triangle formed by gravity, beam, and strut force vectors

We can use trigonometry to solve it.
Because it is a right 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:

Balcony diagram labeled with calculated forces: weight 785 Newtons, beam 1360 Newtons, strut 1570 Newtons

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)?

Person standing on a cantilever beam protruding from a wall

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

The Free Body Diagram looks like this:

Free body diagram of cantilever beam showing downward weight, upward reaction, and a turning moment

The upwards force R balances the downwards Weight.

With only those two forces the beam will spin like a propeller! But there's 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 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.

Person holding a bent fishing rod, experiencing a rotational force or moment

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

Box on a twenty-degree ramp showing gravity, friction, and normal reaction forces

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 isn't accelerating, so the forces are in balance:

Vector triangle of forces acting on the box on a ramp

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:

Vector triangle with solved forces: gravity 981 Newtons, friction 336 Newtons, normal 922 Newtons

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 represented in polar coordinates with magnitude r and angle theta <=> Vector represented in Cartesian coordinates with x and y components
Vector a in Polar
Coordinates
  Vector a in Cartesian
Coordinates

You can read how to convert them at Polar and Cartesian Coordinates, but here's 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!

Top-down view of Sam and Alex pulling a box with ropes at different angles

Example: Pulling a Box

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

  • Sam pulls with 200 Newtons of force at 60°
  • Alex pulls with 120 Newtons of force at 45° as shown

What's the combined force, and its direction?

Let's add the two vectors head to tail:

Two force vectors, 200 Newtons and 120 Newtons, drawn head-to-tail

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

Sam's Vector:

  • x = r × cos( θ ) = 200 × cos(60°) = 200 × 0.5 = 100
  • y = r × sin( θ ) = 200 × sin(60°) = 200 × 0.8660 = 173.21

Alex's Vector:

  • x = r × cos( θ ) = 120 × cos(−45°) = 120 × 0.7071 = 84.85
  • y = r × sin( θ ) = 120 × sin(−45°) = 120 × -0.7071 = −84.85

Now we have:

Force vectors resolved into horizontal and vertical 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:

  • r = √ ( x2 + y2 ) = √ ( 184.852 + 88.362 ) = 204.88
  • θ = tan-1 ( y / x ) = tan-1 ( 88.36 / 184.85 ) = 25.5°

And we have this (rounded) result:
Resultant vector triangle showing a combined force of 204.88 Newtons at 25.5 degrees

And it looks like this for Sam and Alex:
Top-down view showing the single combined force vector pulling the box

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

Elevator

And here's an example using Newton's Second Law:

F = ma
Elevator connected by a cable over a pulley to a counterweight

Example: Elevator Ride

You are in a free-running elevator held up by a weight on a pulley.

  • You and the elevator together weigh 150kg
  • The weight is 125kg

Are you in danger? What will your acceleration be?

Let's start with F = ma

We want to know acceleration so let's rearrange that into:

a = Fm

The total mass that's being accelerated is 150 kg + 125 kg = 275 kg

The force, though, is created only by the difference between weights:

Weight difference = 150 kg − 125 kg = 25 kg

With gravity (g=9.8 m/s2) that makes a force of:

Force = 25 kg × 9.8 m/s2 = 245 N

Let's plug our values in!

a = Fm = 245 N275 kg = 0.89 m/s2

So less than a tenth of gravity's acceleration, just don't fall too far!

(What if the weight was 160 kg?, or 200 kg?)

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