This page covers the following topics:
1. Free-body diagrams
2. Calculating resultant force
3. Resolving forces as vectors
A free-body force diagram is a diagram which shows the relative magnitude and direction of all the forces acting on an object. The object is drawn as a dot and all forces acting on it are drawn as arrows pointing away from it. Each force must be labelled with the name of the force. Both normal reaction force and friction arise from a surface, thus their vectors start from where an object touches the surface. Weight is caused by the mass of the object, thus an arrow representing weight starts at the centre of mass.
Forces can be split into two or more vectors as long as they add up to the original vector. To find the resultant force between two vectors, we can draw the two vectors one after the other, that is the second starts where the first ends. To get an accurate answer, the lengths of the vectors have to be proportional to the sizes of the forces. The resultant force is the vector that joins the start and the finish points. You can determine whether an object is in equilibrium, by drawing all the forces acting on it one after the other. If the last vector ends where the first started, then the object is in equilibrium.
A resultant force is the force obtained when two or more forces are acting on a body. To find the resultant force in one direction, sum forces acting in the same direction and subtract forces acting in an opposite direction. When forces are acting perpendicularly to each other, the Pythagoras' theorem can be used to find the resultant force.
Vectors, including forces, can be resolved into their horizontal and vertical components, usually given as Fₓ and Fᵧ. Their magnitude and direction are usually found using trigonometric functions. The resultant force between forces that are not acting on a straight line or at right angles can be found by resolving them into their horizontal and vertical components, and continuing as usual.
Problems involving objects on slopes can be solved by resolving forces perpendicularly and parallelly to the slope, rather than horizontally and vertically. The angle the slope makes with the horizontal is equal to the angle the weight of the object makes with the line perpendicular to the slope.
A ball is being pulled towards the left across a table as in the free-body diagram provided. Name forces A, B, C, D.
Horizontally towards the left there is going to be a pull (D) and towards the right - friction (B). Vertically the weight of the ball (C) is going to be balanced by the normal reaction force (A).
A = normal reaction force
B = friction
C = weight
D = pull
Jane tries to push a box forward horizontally by exerting a force of 45 N on it. The box does not move, thus she calls her friend, Mary, to help her. Jane pushes again with a force of 45 N and Mary pushes with a force of 38 N, both horizontally. If the frictional force exerted on the box by the ground is 47 N, what is the magnitude of the resultant force acting on the box?
The direction in which they are pushing can be considered to be positive.
resultant force = 45 N + 38 N − 47 N = 36 N in the direction they are pushing in
A mass is placed on an inclined plane. Provide names and the directions of the forces that would you represent in a free-body diagram for the mass.
Normal reaction force would be acting perpendicular to the inclined plane. Friction arising from the surface wold be acting parallel to the surface. Weight would be acting on the climber downwards.
normal reaction force perpendicular to the surface, friction parallel to the surface, downwards weight
Resolve a vector of magnitude 115 N acting 25° above the horizontal into its horizontal and vertical components.
horizontal component Fₓ = 115 × cos(25°) = 104 N (3 s. f.)
vertical component Fᵧ = 115 × sin(25°) = 48.6 N (3 s. f.)
horizontally 104 N, vertically 48.6 N
The diagram provided shows a free-body force diagram of an object. Calculate the magnitude of the resultant force acting on the object.
resolving vertically: 5 N − 3 N = 2 N
resolving horizontally: 8 N − 6 N = 2 N
magnitude of resultant force = √(2² + 2²) = 2.8 N (2 s. f.)
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