This page covers the following topics:
1. Tension and friction
2. Free-body diagrams
3. Calculating resultant force
4. Resolving forces as vectors
Tension is a force that commonly exists in ropes and pulls objects in the direction of the side the rope is attached to. If tension exists within a rope, then any objects attached to it experience tension of the same magnitude and opposite direction.
The friction force is the force which opposes the direction of motion exerted by a surface onto an object moving or trying to move across it. The maximum value of the friction force μR, where μ is the coefficient of friction and R is the normal force. This maximum value is reached when the body it is acting on is in limiting equilibrium, that is, about to move.
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 box of mass 6 kg is lying on rough horizontal surface and is attached to a string which passes over a smooth pulley. The coefficient of friction between the box and the surface is 0.53. A rock is suspended from the other end of the string. Given that the box is in limiting equilibrium, calculate the mass of the rock.
Since the rock is in vertical equilibrium, the normal reaction force is equal to the weight.
R = 6 × 9.8 = 58.8 N
F = μR
friction = 0.53 × 58.8 N = 8.374 N
Since the box is in horizontal equilibrium, tension = friction = 8.374 N. Since the pulley is smooth, the tension on both sides of it is equal. Solving for the rock, since it is in equilibrium, tension = weight.
w = mg
8.374 N = mass × 9.8 m/s²
mass = 3.18 kg
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
In a space station a 3 kg box is being held at equilibrium suspended by two ropes attached on opposite sides. One of the ropes has a tension of 15 N. Find the tension within the second rope.
According to Newton’s first law, since the box is in equilibrium and in the space station the weight can be neglected, the two tension forces must be equal to 15 N.
A painting is suspended in equilibrium by two inextensible vertical strings. Given that the tension in each of the strings is 49 N, find the mass of the painting.
According to Newton’s first law, since the painting is in equilibrium, tension forces balance the weight.
w = 2T
w = 2 × 49 N = 98 N
w = mg
98 N = m × 9.8 m/s²
mass = 98 ÷ 9.8 m/s² = 10 kg
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
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