# Calculating the cord tension from a ball moving in a circle

I've been working on a basic circular motion exercise from the book about a ball moving around some circle attached to a cord - but I don't understand the way the book is calculating the tension of the cord. Here is what I did:

Have a small ball of $0.5$kg at the end of a cord. It swings in a section of a vertical circle of radius $2$m at a speed of $8$m/s.

What is the tension of the string when the angle is $20$?

Well, first we note that the weight of the ball is $0.5*9.8=4.9$N:

We can draw a right triangle here and find the hypotenuse, which is the ball is applying away from the cord:

The cosine of $20$ is $\frac{4.9}{hypotenuse}$ so the hypotenuse is

$$\frac{4.9}{\cos20} = 5.21$$

Additionally we have to find the radial acceleration, which comes simply from

$$\frac{m\cdot v^2}{r} = \frac{0.5 \cdot 8^2}{2} = 16$$

So, the tension of the cord is the centrifugal force $16$ added by the $5.21$ caused by the weight of the ball.

However, the book shows that instead of $5.21$ it should've been $4.6$, which comes from

$$4.9 \cos 20 = 4.6$$

But I don't understand why would they do that. $\cos 20$ is $\frac{4.9}{hypotenuse}$, so multiplying it by $4.9$ should not yield anything meaningful. On the other hand, you can get the hypotenuse just by dividing $4.9$ by $\cos 20$.

What did I overlook?

• The component of a force $F$ in a direction making an angle $\theta$ with it is $F\cos\theta$ - not $\dfrac{F}{\cos\theta}$. Think about it. How can the influence of a force be greater than the force itself in a direction different from the direction in which it is acting? This is not a complete answer or a proof of how the $F\cos\theta$ comes but this is an argument that can intuitively motivate you to believe that $\dfrac{F}{\cos\theta}$ is the wrong expression. – Dvij Mankad Nov 5 '17 at 11:15
• You are wrong in interpreting the diagrams wrongly. Here is a link that might help you understand how to resolve vectors (i.e. force) into components. physicsclassroom.com/class/vectors/Lesson-1/Vector-Resolution – Dvij Mankad Nov 5 '17 at 11:17
• Another thing: Your question is simply about why the component of the force along the cord is not $\dfrac{F}{\cos\theta}$ and why it is $F\cos\theta$. You might want to shorten the question to increase the readability by cutting out irrelevant details. – Dvij Mankad Nov 5 '17 at 11:19
• @Dvij but I am doing exactly what they are - in fact, if you take their example image and rotate it such that it fits my case (i.imgur.com/xLkSrfw.png) you can get the hypotenuse (60) by performing $$\frac{45.9}{\cos 40} = 60$$ Why would my case be any different? – Omega Nov 6 '17 at 1:26