The resultant force of two equal parallel forces acting in opposite direction in same line of action

The forces, which are having their line of action parallel to each other, are known parallel forces. The two parallel forces will not intersect at a point.

The following are the important types of parallel forces :

1. Like parallel forces,

2. Unlike parallel forces.

1. Like parallel forces. The parallel forces which are acting in the same direction, are known as like parallel forc.es. In Fig. 1.58, two parallel forces F1 and F2 are shown. They are acting in the same direction. Hence they are called as like parallel forces. These forces may be equal or unequal in magnitude.

Unlike Parallel Forces:

The parallel forces which are acting in the opposite direction, are known as unlike parallel forces. In Fig. 1.59, two parallel forces F1 F2 are acting in opposite direction. Hence they are called as unlike parallel forces. These forces may be equal or unequal in magnitude.

The unlike parallel forces may be divided into : (i) unlike equal parallel forces, and (ii) unlike unequal parallel forces.

Unlike equal parallel forces are those which are acting in opposite direction and are equal in magnitude.

Unlike unequal parallel forces are those which are acting in opposite direction and are unequal in magnitude.

The resultant of following two parallel forces will be considered:

1. Two parallel forces are like.

2. Two parallel forces are unlike and are unequal in magnitude.

3. Two parallel forces are unlike but equal in magnitude.

Resultant of Two Like Parallel Forces:

Fig. 1.60 shows a body on which two like parallel forcesF1 andF2 are acting. It is required to determine the resultant (R) and also the point at which the resultant R is acting. For the two parallel forces which are acting in the same direction, obviously the resultant R is given by,

R =F1 +F2

In order to find the point at which the resultant is acting, F Varignon’s principle (or   method of moments) is used. According 1 60 IQ to this, the algebraic sum of moments of F 1 and F2 about any point should be equal to the moment of the resultant (R) about that point. Now arbitrarily choose any point 0 along line AB and take moments of all forces about this point.

Moment of F 1 about 0 = F 1 x AO (clockwise)(-)

Moment of F2 about 0 = F2 x BO (anti-clockwise)(+ ve)

. . Algebraic sum of moments of F1 and F2 about O= – F 1 x AO + F 2 x BO

Moment of resultant about 0 = Rx OC (anti-clockwise){+)

But according to principle of moments the algebraic sum of moments of F 1 and F2 about

0 should be equal to the moment of resultant about the same point 0.

F 1 x AO + F2 x BO = +R x CO= (F1 + F2) x CO

or .                                F1(AO +CO)= F2(BO- CO)

F l x AC = F 2 x BC

F1 / F2 = BC / AC

The above relation shows that the resultant R acts at the pointe, parallel to the lines of action of the given forcesF1 andF2 in such a way that the resultant divides the distance AB in the ratio inversely proportional to the magnitudes of F 1 and F2 Also the point C lies in line AE i.e., point C is not outside AB.

The location of the pointe, at which the resultant R is acting, can also be determined by taking moments about points A of Fig. 1.60. As the forceF1 is passing through A, the moment of F 1 about A will be zero.

The moment of F 2 about A= F2 x AB (anti-clockwise)(+)

Algebraic sum of moments of F 1 and F2 about 0

= 0 + F2 x AB = F2 x AB (anti-clockwise) (+)

The moment of resultant R about A

= R x AC (anti-clockwise)(+)

But according to the principle of moments, the algebraic sum of moments of F1 and F 2

about A should be equal to the moment of resultant about the same point A. Hence equating

equations (i) and (ii),

F2 x AB =R x AC

But R = (F 1 + F2 ) hence the distance AC should be less than AB. Or in other words, the

point C will lie inside AB.

Resultant of Two Unlike Parallel Forces (unequal in magnitude):

Fig 1.61 shows

a body on which two unlike parallel forces F 1 and F2 are acting which are unequal in magnitude. Let us assume that forceF1 is more than F2• It is required to determined the resultant and also the point at which the resultant R is acting. For the two parallel forces, which are acting in opposite direction, obviously the resultant is given by,

R=F1 -F2

Let the resultant R is acting at C as shown in Fig. 1.61.

In order to find the point C, at which the resultant is acting, principle of moments is used.

Choose arbitrarily any point 0 in line AB. Take the moments of all forces (i.e., Fp F2 and R) about this point.

Moment of F 1 about 0 = F 1 x AO (clockwise)

Moment of F2 about 0 = F2 x BO (clockwise)

Algebraic sum of moments of F 1 and F2 about 0

=F1 x AO +F2 x BO

Moment of resultant force R about 0

= R x CO (clockwise)

= (F1 – F2) x CO

= F1 x CO –F2 x CO

But according to the principle of moments, The algebraic sum of moments of all forces about any point should be equal to the moment of resultant about that point. Hence equating equations (i) and (ii), we get

F 1 x AO + F2 x BO = F 1 x CO – F 2 x co

F2(BO +CO) = F 1(CO -AO)

F2 x BC=F1 x AC

BC / AC = F1 / F2 or F1 / F2 = BC / AC

But F1 >F2, hence BC will be more than AC. Hence point O lies outside of AB and on the same side as the larger force F1 Thus in case of two unlike parallel forces the resultant lies de the line joining the points of action of the two forces and on the same side as the larger

The location of the Point C, at which the resultant R is acting, can also be determined by taking moment about point A, of F1g. 1.61. As the force F1 is passing through A, the moment of F1 about A will be zero.

The moment of F2 about A= F2 x AB (clockwise)(-)

Algebraic sum of moments of F 1 and F2 about A

= O + F2 x AB = F2 x AB (clockwise)(- ) … (i)

The moment of resultant R about A should be equal to the algebraic sum of moments of F1 and F2 (i.e.,= F2 x AB) according to the principle of moments. Also the moment of resultant R about A should be clockwise. As R is acting upwards [ ·: F1 > F2 and R = (F1 – F2) so R is acting in the direction ofF1], the moment of resultant R about A would be clockwise only if the points C is towards the left of point A. Hence the point C will be outside the line AE and on the side of F 1 (i.e. , larger force).

Now the moment of resultant R about A

= R x AC (clockwise)(-) … (ii)

Equating equations (i) and (ii),

F2 x AB =R x AC

= (F1 -F2) x AC (·: R=F1 – F2)

As F1, F2 and AB are known, hence AC can be calculated. Orin other words, the location of point C is known.

Resultant of Two Unlike Parallel Forces which are Equal in Magnitude:

When two equal and opposite parallel forces act on a body, at some distance apart, the two forces from a couple which has a tendency to rotate the body. The perpendicular distance between the parallel forces is known as arm of the couple.

Fig. 1.62 shows a body on which two parallel forces, which are acting in opposite direction but equal in magnitude are acting. These two forces will form a couple which will have a tendency to rotate the body in clockwise direction. The moment of the couple is the product of either one of the forces and perpendicular distance between the forces.

Let       F =Force at A or at B

a= Perpendicular distance (or arm of the couple)

The moment (M) of the couple is given by, M = F x a.

The units of moment will be Nm.

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In this article, you will learn what the resultant force (also known as net force) is, and how to find it when an object is subject to parallel forces as well as non-parallel forces with the help of examples.

What is the resultant force?

When an object is subject to several forces, the resultant force is the force that alone produces the same acceleration as all those forces.

For example, if 4 forces act on a block and cause it to accelerate 1 m/s2 south, then the resultant force is the force that, if applied alone to the block, will also make it accelerate 1 m/s2 south.

The reason why the resultant force is useful is that it allows us to think about several forces as though they were a single force. This means that to determine the effect that several forces have on an object, we only need to determine the effect that a single force has.

How to find the resultant force?

If we know the mass m of an object and the acceleration a produced by the forces that act on it, we can find the resultant force using Newton's Second Law. Indeed, according to Newton's Second Law, the force F that alone produces the acceleration a on an object of mass m is:

This force F is our resultant force. So, we can write:

Which indicates that the resultant force R has the same direction as a, and has magnitude equal to the product ma.

For example, if a box of 1.5 kg is subject to 5 forces which make it accelerate 2.0 m/s2 north-west, then the resultant force is directed north-west and has the magnitude equal to 1.5 kg × 2.0 m/s2 = 3.0 N.

Often, however, we know the forces that act on an object and we need to find the resultant force.

Experiments show that when an object is subject to several forces, F1, F2, ..., the resultant force R is the vector sum of those forces:

Notice that this is not a mere sum of the magnitudes of the forces, but the sum of the forces taken as vectors, which is more involved because vectors have both a magnitude and a direction that we need to consider when doing the sum.

According to the above equation, if an object is subject to no forces, then the resultant force is zero, and if an object is subject to only one force, then the resultant force is equal to that force. These two cases are pretty simple, but what about an object subject to two or more forces? How do we perform the vector sum then?

To explain this clearly, we will now go through all the cases that can happen, from simple ones in which all the forces are parallel, to more complex ones in which the forces are not parallel, and show how to find the resultant force in each of them with the help of examples.

• Two forces acting in the same direction

Let's start with the simple case in which an object is subject to two forces that act in the same direction:

The resultant force is in the same direction as the two forces, and has the magnitude equal to the sum of the two magnitudes:

• Two forces acting in opposite directions

Let's consider the case in which an object is subject to two forces that act in opposite directions.

If the two forces are equal in magnitude:

The resultant force will be zero because two opposite forces cancel each other out.

On the other hand, if the two forces are not equal in magnitude:

The resultant force will be in the same direction as the force with the larger magnitude (the 5 N force in the example), and have the magnitude equal to the difference between the magnitudes of the two forces (in the example that would be 2 N):

• More than 2 forces parallel to one another

Let's now consider the case in which an object is subject to more than two parallel forces:

To find the resultant force in this case, we first sum all the forces that go in one direction, and then all the forces that go in the other direction:

At this point, we have two forces that are in opposite directions, which is a case that we already know how to solve: the resultant force has the same direction as the force with the larger magnitude (the 11 N force), and its magnitude is equal to the difference between the two magnitudes (4 N):

• Two forces that are not parallel

In the previous cases, we have forces that are all parallel to one another. It's time to consider the case in which an object is subject to two forces that are not parallel.

For example, let's assume that we have a block subject to two forces, F1 and F2.

F1 has magnitude 50 N and is applied at a 45° angle, whereas F2 has magnitude 60 N and is applied horizontally, as shown in the free-body diagram below:

How do we find the resultant force R in this case?

The first step is to draw coordinate axes on our free-body diagram.

Since one of the two forces is horizontal, for convenience, we choose the x-axis horizontal, and the y-axis vertical, and we place the origin at the center of our block:

The next step is to determine the x and y components of all the forces that act on the block:

Now comes the important part:

If we sum all the x components, we will get the x component of the resultant force:

F1x + F2x = Rx

Rx = F1x + F2x

Rx = F1 cos 45° + F2

Rx = (50 N) (cos 45°) + 60 N

Rx = 95 N

Similarly, if we sum all the y components, we will get the y component of the resultant force:

F1y + F2y = Ry

Ry = F1y + F2y

Ry = F1 sin 45° + 0

Ry = F1 sin 45°

Ry = (50 N) (sin 45°)

Ry = 35 N

At this point, we know the x and y components of R, which we can use to find the magnitude and direction of R:

The magnitude of R can be calculated by applying Pythagoras' Theorem:

R = √Rx2 + Ry2

R = √952 + 352 N = 100 N

The angle θ that R makes with Rx can be calculated using trigonometry:

Thus, the resultant force R has magnitude 100 N and direction angle of 20°.

• More than 2 non-parallel forces

Finally, let's examine the case in which an object is subject to more than two non-parallel forces.

For example, suppose we have an object that is subject to three forces, F1, F2, and F3.

The magnitude of each force is shown below:

F1 = 10 N

F2 = 20 N

F3 = 40 N

The free-body diagram of the object looks like this:

We can find the resultant force R using the same process that we used in the previous case of two non-parallel forces.

First, we draw the coordinate axes on our free-body diagram:

Then, we determine the x and y components of the individual forces:

F1x = F1

F2x = 0

F3x = −F3 cos 60°

F1y = 0

F2y = F2

F3y = −F3 sin 60°

Again, the x component of the resultant force R is the sum of all x components:

Rx = F1x + F2x + F3x

Rx = F1 + 0 + (−F3 cos 60°)

Rx = F1 − F3 cos 60°

Rx = 10 N − (40 N) (cos 60°)

Rx = −10 N

Similarly, the y component of R is the sum of all y components:

Ry = F1y + F2y + F3y

Ry = 0 + F2 + (−F3 sin 60°)

Ry = F2 − F3 sin 60°

Ry = 20 N − (40 N) (sin 60°)

Ry = −15 N

Finally, let's calculate the magnitude and direction of R using its two components Rx and Ry:

R = √Rx2 + Ry2

R = √(−10)2 + (−15)2 N = 18 N

To express the direction of R, we need to calculate the direction angle (i.e. the counterclockwise angle that R makes with the positive x-axis), which in our case is 180° + θ, i.e. 236°.

The process that we used in this case and in the previous one to find the resultant force when the forces are not parallel can also be used when all the forces are parallel. In fact, it can be used in any case – it's a generic process. However, in the cases of parallel forces, we recommend using the much simpler processes that we described before.

Here's a quick summary of the generic process:

A note on drawing coordinate axes on a free-body diagram: we recommend you to draw them so that one of the axes is in the same direction as the acceleration of the object. For example, if you have an object accelerating up a ramp, you should draw tilted coordinate axes with the x-axis uphill. Sometimes, however, your object may be at rest or you may not know the direction of the acceleration. In that case, place the coordinate axes so that as many forces as possible are parallel to them since this will simplify the expressions for their components.

To test your understanding, do the exercises below.

Exercises

#1

John and Rob are engaged in a tug of war. John is pulling with a force of 230 N, and Rob is pulling with a force of 215 N. Determine the magnitude and direction of the resultant force.

#2

A car of 1400 kg is subject to multiple forces which produce an acceleration of 3.5 m/s2 directed north. Find the net force.

The direction of Fnet is the same as that of a (north), and the magnitude is:

Fnet = ma

Fnet = (1400 kg) (3.5 m/s2)

Fnet = 4900 N

#3

A block is pulled by two forces of 15 N and 25 N to the left, and by three forces of 10 N, 20 N, 30 N to the right. Find the magnitude and direction of the resultant force.

If you sum the forces pulling to the left, you get 40 N to the left, and if you sum the forces pulling to the right, you get 60 N to the right.

Thus, the resultant force is 20 N to the right.

#4

An apple is subject to two vertical forces: one of 40 N pulling upward, and the other of 10 N pulling downward. What is the net force acting on the apple?

#5

A box of 1.0 kg is in free fall (i.e. moving subject only to the force of gravity). Calculate the net force.

The net force is equal to the force of gravity because the box is subject only to that one force. Therefore, the direction is downward, and the magnitude is:

Fnet = mg

Fnet = (1.0 kg) (9.8 m/s2)

Fnet = 9.8 N

#6

A tugboat is horizontally pulled by two forces of 1450 N, each making an angle of 20° with the long axis of the tugboat, as shown in the figure (the view is from the above):

Assuming there is no friction, what is the magnitude and direction of the resultant force acting on the tugboat?

#7

A ball is subject to two forces F1 and F2. The magnitudes of the two forces are 45.0 N and 70.0 N respectively. In the figure below you can see the free-body diagram of the ball:

Find the magnitude and direction of the resultant force acting on the ball.

#8

An empty box is pulled by two men with horizontal forces, as shown below (the view is from the above):

Assuming that F1 is 345 N and F2 is 458 N, and there is no other horizontal force acting on the box, find the magnitude and direction of the resultant force.

Problems with solutions

To further test your understanding of resultant forces, see our force problems, which include problems where you need to find the resultant force acting on objects that move horizontally, move up an incline, and hang from pulleys. For each problem, we provide a step-by-step guide on how to solve it.