Balanced Forces Resolving and finding the resultant force...

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Balanced Forces Resolving and finding the resultant force...

Learning Objective : Balanced Forces To start to use our vector adding and resolving skills to solve systems of balanced forces Book Reference : Pages 94-96

Balanced Forces Firstly a starter question: A point object with a weight of 6.2N is acted upon by a horizontal force of 3.8N. Calculate the resultant of these two forces Calculate the magnitude and direction of the 3rd force which keeps the object balanced

First draw a picture... Horizontal force of 3.8N θ Weight 6.2N R combined resultant force

Our two original forces can be replaced by our single resultant force To find the magnitude of the resultant vector we use Pythagoras’: R = √((3.2)2 + (6.2)2 )= 7.3N and we use trig’ to find the direction tan θ = O/A = 3.8/6.2 θ = 31.5° to the vertical Horizontal force of 3.8N θ Weight 6.2N R resultant Our two original forces can be replaced by our single resultant force

For our system of forces to be balanced, the balancing force must be equal and opposite 7.3N at 31.5° to the vertical θ θ R resultant

What is the resultant of the following system of forces? Hint find a suitable pair of perpendicular forces and resolve all forces in those directions before finding the resultant Answer 19.72N at 4.57° to the 10N force

We have a suitable pair of perpendicular forces, (8N and 12N) find the components of the 10N force in these directions Resolving to find the component of the 10N force in the direction of the 12N force 10 cos 40° = 7.66N Therefore total forces in this direction : 12N + 7.66N = 19.66N 40°

Resolving to find the component of the 10N force in the direction of the 8N force 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 40° 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

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° θ° 19.66N We can say that the resultant force is 19.72N at 4.57° left to right to the original 10N force

Summary : We’ve seen that we can replace systems of forces with a single resultant force. We achieved this in two steps: 1. Resolving all forces in the direction of a pair of perpendicular forces to yield only a pair of equivalent perpendicular forces. 2. Finding the equivalent single resultant force which can replace the pair of perpendicular forces

Summary 2 : Often the scenarios you will be asked to solve will be a collection of balanced forces in equilibrium. This tells you that the resultant force is zero... For example : a stationary rough object on a rough inclined plane. The friction is holding the object in place. The system is balanced or in equilibrium