Key Areas covered

Key Areas covered • Resolving a force into two perpendicular components. • Forces acting at an angle to the direction of movement.

What we will do today: • Identify how to solve problems when we have two forces acting at different angles (component of forces). • Carry out problems on the above

Resolution (Rectangular Components) of Forces

Any vector, x, can be resolved into two components at right angles to each other. The horizontal component xh The vertical component xv F xv is equivalent to θ xh

sin θ = xv / x xv = x sin θ cos θ = xh / x xh = x cos θ x xv θ xh

Force The vertical and horizontal components of a Force vector, F, are, respectively: Fv = F sin θ Fh = F cos θ

Components of Forces

In the previous section, a vector was split into horizontal and vertical components. This can obviously apply to a force. is equivalent to Remember that the resultant of a number of forces is the single force which has the same effect, in both magnitude and direction, as the sum of the individual forces. Fv = F sin θ F θ Fh = F cos θ

A man pulls a garden roller of mass 100 kg with a force of 200 N acting at 30º to the horizontal. If there is a frictional force of 100 N between the roller and the ground, what is the acceleration of the roller along the ground? Solution Fh = F cos θ = 200 cos 30º = 173.2 N Fun = 173.2 N – Friction = 173.2 – 100 = 73.2 N a = Fun / m = 73.2 / 100 = 0.732 ms-2 Example 1 200 N 30º Fh Friction = 100 N

Example 2 • An oil rig is pulled by two barges as shown. Each barge applies a force of 10 kN. What is the resultant force acting on the rig? Solution: Work out Fh for one barge: Fh = F cos θ = 10 000 cos 30 = 8660N Then double this to include forces from both barges: 2 x 8660 = 17 300 N (3 sig fig)

Experiment

2004 Qu: 1

2009 Qu: 3

2001 Qu: 21

Questions • Activity sheets: • Page 31 – 33 • Qu’s 1-5 • Resolution of forces • You should now be able to answer all questions in class jotter

Key Areas covered