Learning Objective :Learning Objective :

1.1. To To startstart to use our vector adding and to use our vector adding and resolving skills to solve systems of resolving skills to solve systems of balanced forcesbalanced forces

Book Reference : Pages 94-96Book Reference : Pages 94-96

Firstly a starter question:Firstly a starter question:

A point object with a weight of 6.2N is A point object with a weight of 6.2N is acted upon by a horizontal force of 3.8N.acted upon by a horizontal force of 3.8N.

a)a)Calculate the resultant of these two Calculate the resultant of these two forcesforcesb)b)Calculate the magnitude and direction of Calculate the magnitude and direction of the 3the 3rdrd force which keeps the object force which keeps the object balancedbalanced

R combined resultant force

θ

Horizontal force of 3.8N

Weight 6.2N

R resultant

θ

Horizontal force of 3.8N

Weight 6.2N

To find the magnitude of the To find the magnitude of the resultant vector we use Pythagoras’:resultant vector we use Pythagoras’:

R = √((3.2)R = √((3.2)22 + (6.2) + (6.2)2 2 )= )= 7.3N7.3N

and we use trig’ to find the directionand we use trig’ to find the direction

tan tan θθ = O/A = 3.8/6.2 = O/A = 3.8/6.2

θθ = 31.5° to the vertical = 31.5° to the vertical

R resultant

θ

θ

What is the resultant of the following system of What is the resultant of the following system of forces?forces?

40o

90o8 N

10 N

12 N

Hint find a suitable pair of perpendicular forces Hint find a suitable pair of perpendicular forces and and resolveresolve all forces in those directions before all forces in those directions before finding the finding the resultantresultant

Answer 19.72N at 4.57° to the 10N forceAnswer 19.72N at 4.57° to the 10N force

We have a suitable pair of perpendicular We have a suitable pair of perpendicular forces, (8N and 12N) find the components forces, (8N and 12N) find the components of the 10N force in these directionsof the 10N force in these directions

ResolvingResolving to find the component of the to find the component of the 10N force in the direction of the 12N force10N force in the direction of the 12N force

40°

10 cos 40° = 7.66N

Therefore total forces in this direction :

12N + 7.66N = 19.66N

Resolving to find the component of the Resolving to find the component of the 10N force in the direction of the 8N force10N force in the direction of the 8N force

40°

10 sin 40° = 6.43N

But this is acting in the opposite direction to the 8N force

Therefore total forces in this direction :

8N - 6.43N = 1.57N

We have now collapsed all of the forces in the original system to a simple pair of perpendicular forces and we know how to find the resultant force in these circumstances

θ°

1.57N

19.66N

R

Using Pythagoras and trigonometry we can find the single resultant force

To find the magnitude of R

R = √((1.57)2 + (19.66)2) = 19.72N

To find the direction of R

Tan θ = O/A = 1.57/19.66

θ = 4.57°We can say that the resultant force is 19.72N at 4.57° left to right to the original 10N force