1.1 - Resultant Force

Meaning and Direction

Resultant force is one force equivalent to the vector sum of all forces acting on an object.
Equal forces in opposite directions give zero resultant force; an initially stationary object remains stationary.
Unequal opposing forces give a resultant in the direction of the larger force. In tug-of-war, the initially stationary rope moves towards the stronger pull.
State both the magnitude and direction of a resultant.

Arrangement	Method
Same direction	Add magnitudes; retain their direction. Forces of 6 N and 8 N to the right give 14 N to the right.
Opposite directions	Use signed addition. Taking right as positive, $- 8 + 6 = - 2 N$ means 2 N to the left.
Perpendicular	Treat the forces as adjacent sides of a rectangle; its diagonal gives the resultant.
Neither parallel nor perpendicular	Use a scaled triangle of forces or parallelogram of forces.

For perpendicular forces of 8 N horizontally right and 6 N vertically up:

F = 8^{2} + 6^{2} = 10 N, tan θ = \frac{6}{8}, θ = 36. 9^{\circ}

Triangle method:

Choose a scale and draw the first vector.
Draw the second vector from the first vector’s arrowhead, preserving its direction.
Join the first vector’s tail to the second vector’s head to obtain the resultant.
Measure the resultant’s length and convert using the scale; measure its direction with a protractor.

Parallelogram method:

Choose a scale and draw both vectors from a common point.
Complete the parallelogram using a pair of compasses.
Draw the diagonal from the common point to represent the resultant.
Measure the diagonal’s length and convert using the scale; measure its direction with a protractor.

Strings over pulleys apply forces to a metal ring. A third force can balance the resultant of the other two.
When the ring is centred without touching the metal rod, the force through pulley C balances the resultant of the forces through pulleys A and B.
The balancing force has the same magnitude as that resultant and acts in the opposite direction. This allows the resultant’s magnitude and direction to be determined experimentally.

A free body diagram shows all forces acting on the selected object only. A book’s diagram includes its weight and the table’s normal reaction; it excludes the force the book exerts on the table.
A bag supported by a wedge on an inclined plane experiences weight vertically downwards, the plane’s normal reaction perpendicular to the plane and the wedge’s normal reaction up the plane.
A trailer experiences weight, normal reaction, pulling force and friction. A rising rocket experiences engine thrust upwards, with weight and air resistance downwards.
Newton’s Second Law of Motion:

F = ma

Here $F$ is the resultant force, in N; $m$ is mass, in kg; $a$ is acceleration, in $m s^{- 2}$ .

State of a car on a horizontal road	Force relationships
Stationary	$v = 0$ , $a = 0$ , $F = 0$ ; normal reaction $R$ equals weight $W$ .
Uniform velocity	$a = 0$ , $F = 0$ ; $R = W$ and engine thrust $T$ equals the opposing force $F_{R}$ .
Increasing velocity with uniform acceleration	$a \neq = 0$ , $F \neq = 0$ ; $R = W$ , $T > F_{R}$ and $F = T - F_{R}$ .

Falling coconut: $m = 2.0 kg$ , downward $a = 9.0 m s^{- 2}$ and $g = 9.81 m s^{- 2}$ . Weight acts downwards and air resistance $R$ upwards.

F = ma = 2.0 (9.0) = 18.0 N downwards

W = m g = 19.62 N, W - R = F, R = 19.62 - 18.0 = 1.62 N

Passenger in a lift: $m = 60 kg$ , $g = 9.81 m s^{- 2}$ . The floor’s normal reaction $R$ acts upwards and weight $W = 588.6 N$ downwards.
At rest or at uniform velocity $8.0 m s^{- 1}$ , $F = 0$ , so $R = W = 588.6 N$ .
With upward acceleration $1.2 m s^{- 2}$ :

R - W = ma, R = 588.6 + 60 (1.2) = 660.6 N

A 1.2 kg trolley is pulled along a table by a hanging load; acceleration is $4.0 m s^{- 2}$ and opposing friction is 6.0 N.
On the trolley, weight and normal reaction balance vertically. Tension $T$ pulls towards the pulley; friction $F_{R}$ opposes motion.
On the load, weight $B$ acts downwards and tension $T$ upwards. The connected trolley and load have the same acceleration.

F_{trolley} = 1.2 (4.0) = 4.8 N

T - F_{R} = 4.8, T = 4.8 + 6.0 = 10.8 N

B - T = ma, 9.81 m - 10.8 = 4.0 m

m = \frac{10.8}{9.81 - 4.0} = 1.86 kg