In this article, you’ll be able to review how to solve force equations which is a subject you will come across in the mechanics' math module. This is a subject that can be hard to understand at once - but don´t worry, we’ll go through the concepts one by one. You’ll need a scientific calculator that will help you with half of the work, and a lot of energy from you to help beat the stress.

We’ll start by looking at core concepts that we'll need to be familiar with, before moving onto some trickier problems. Before getting started, be sure you have a clear understanding of algebra.

You’ll find more information to solve Differentiation/integration problems in this link.

Gravity (heading 2)

We’ll start by looking at some key facts about gravity. You should know that gravity's acceleration is 9.8ms^-2 on Earth and that it has a symbol to substitute this figure, which is the letter "g".

You will also have to freshen up your mind a bit with physics. You may remember that a force (Newton) is equal to the mass of an object multiplied by its acceleration and it looks like this: F = ma.

This means that gravity is just an acceleration, and if you tweak the equation above you will end up with this new one:

W = mg

This equation states that weight is equal to the mass of an object, multiplied by the acceleration of gravity.

Confused already? Don’t worry, we’ve created this free resource to help you find a solution to the stress of dealing with AP math. Remember you’ll have to work hard because a calculator won’t do all the job for you.

Trigonometry (heading 2)

Before we get started, you need to remember basic trigonometric functions, for example, SOHCAHTOA. If it doesn’t sound familiar go ahead and look it up, you’ll need it ahead.

This will come in useful while solving a forces problem because you’ll have to split each force up into an x (horizontal), and y (vertical) component, but more on that below.

Resolving forces (heading 2)

Let's start by finding the solution to some force problems! As this is an article designed to help you out, the first question we’ll present you with is theoretical. But rest assured, you'll come across these and other full-body questions throughout your course, and the key procedure is to simplify everything until you are left with a set of numbers that you can solve for.

Another thing to remember: always draw a diagram!

Let's say there are two forces acting on an object. One force acts horizontally, with a magnitude of 50N, while another force acts at an angle of 20 degrees, with a magnitude of 20N. What is the final force acting on this object?

As you can see, these two forces are acting in different directions, so if you add them together you’ll get the final force. By "final" we mean “overall”, or “total”.

Our force of 50N is only acting in one direction (horizontally), which means there's not much to do. We'll be focusing on the force of 20N that's acting at an angle of 20 degrees And as mentioned above, you’ll have to split this one into its component forces. So, the amount it acts horizontally (on the x axis), and the amount it acts vertically (on the y axis). Don't stress, it's not too hard!

Imagine the 20N force acting at 20 degrees as a triangle, which means that you have a right angle triangle - with an acute angle of 20 degrees. Then, the hypotenuse of the triangle would be 20N.

With some simple use of algebra and the help of SOHCAHTOA, you’ll be able to calculate the forces that act on the adjacent and opposite triangle sides. First, you can start by working out the horizontal force, which acts on the x-axis, and is the adjacent side of our triangle.

Using SOHCAHTOA:

CAH, from SOHCAHTOA

Now, you know that the angle is 20 and that the hypotenuse is 20 as well. At this point, you can move somethings around and re-arrange the equation to find the magnitude of our force in the horizontal direction:

You can now replace with numbers:

The force is acting at 18.8 Newtons in the horizontal direction.

The next step is to find out the amount of force acting in the vertical direction, and for that, there is a similar calculation you can use by replacing cos with sin.

From SOHCAHTOA:

The SOH, of SOHCAHTOA

Going back to your algebra knowledge, you can re-arrange this equation to find the opposite side of the triangle. This side of the triangle is the vertical one and represents the force's magnitude on the y axis.

Since you already have the angle (which theta represents), and the hypotenuse, by replacing them you’ll be able to find the force acting vertically:

Great! The force acting vertically from our original 20N force at an angle of 20 degrees is 6.8N!

Learn about math tutors near me.

Putting everything together (heading 2)

At this point, you know that there was a force of 50N acting horizontally on our object, and a force of 20N acting at a 20-degree angle. What you’ve done so far is split that 20N force into a 6.8N vertical force, and an 18.8N vertical force.

To find the result, all you need to do is put all these values together.

Vertical forces (heading 2)

Let’s get started with vertical forces, which can be pretty easy:

6.8N + 0 = 6.8N

As you can see, the 50N horizontal force doesn't act at all in the vertical direction, so the amount of force there is 0. Now, you’ve ended up with the vertical component of the 20N force.

Horizontal forces (heading 3)

This one can be a little more complicated but you still have to continue trusting your intuition:

50N + 18.8N = 68.8N

You can take the 50N horizontal force and add it to the horizontal component of the 20N force. Like this, you’ll end up with a total force of 68.8N on the x-axis.

The final triangle (heading 3)

After all those steps, you’ll see that you have 2 components of the final force, but not the overall force just yet. In other words, you know the force horizontally, and the force vertically, but you haven’t yet combined them, and so you know know how the final force will look like.

To solve this, you will use one last triangle.

Place your horizontal force of 68.8N on the adjacent side of the triangle, and your vertical force of 6.8N on the opposite side of the triangle.

If you’re asking yourself how does one calculate the combined force, you’ll have to use the famous triangular equation... Pythagoras' theorem! With both sides of the triangle, what you’ll have to do to calculate the hypotenuse is solve this equation:

Pythagoras' theorem

Once you replace the values you have in the equation and finish solving it, the force would be:

So there it is, the answer is 69.1N.

Bonus point! (heading 3)

Your teacher might ask you to find the angle of the resultant force. So remember, now that you know the value for all three sides of the triangle, you’ll be able to find the angle easily using SOHCAHTOA.

To find the angle, you need to use the inverse tan function.

Put in words, the answer is that the force acts at an angle of 5.6 degrees.

Summary (heading 2)

Thank you for making it to the end, we hope you found this article helpful. The best thing to do right now is to keep studying. You could check some previous exam questions or go through your homework and worksheets where you were asked to solve the problem we just solved above. An in-depth analysis of previous work always helps.

Keep in mind that revision and practice are the key elements for success, especially in a math class. It might be intimidating at first, but remember that once you understand the principles, each question is the same.

It's true that this subject can be complicated so use your free time to practice and find a system that helps you retain information.

If you’re still feeling uneasy about the subject, we'd recommend finding a maths tutor on Superprof to help you study and figure out how to solve the equations.

Finding outside help, like the one a tutor can provide, will help you grasp the concepts you might have missed in class, and you’d be doing it at your own pace & energy and learning in ways that suit you best.

You'll be a mathematician in no time!

For yet more tricky Maths problems explained, visit these guides on Calculus and Exponentials and Logs.

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