# Why do we write T cos theta = mg for this Q instead of mg cos theta = T

For this Q , the author of the textbook writes that T cos 45 =mg

but it also states that mg cos 45 is not equal to T. I got to know about this because this is the way , I solved it first.

My Q :1 is that why mg cos 45 is not equal to tension or T.

The main point to noice is that Tcos 45 = mg doesn’t give same value of T as T=mg cos 45.

According to me why I think mg cos 45 should be equal to T :

1. There is no acceleration along T. Therefore acc=0.

2. T and mg cos 45 also come at the same vertical line.

Also as Q2: can we say there is $$a_t$$ is along z axis for this question ?

Name of author: Dc pandey

• you are saying that "they say that T cos 45 =mg" and the continue on "but mg cos 45 is not equal to T", then in the following line "Why mg cos 45 = T according to me :". I really don't understand what the solution said, or what you think, or even what the question is. And then on top of that you follow with another question "Also , can we say there is $a_t$ is along z axis for this question ?". This is very confusing and its becoming a commonplace with your questions. – NMech May 21 at 15:57
• @NMech I humbly apologize for the confusions I made in my Q. I have edited it nicely now. – Srijan M.T May 21 at 17:07
• my apologies for being so blunt. Its nice to see you so eager to learn new things, so personally I am equally eager to give you my best, so please try to form better your questions. For example, in "For this Q , Textbook write that T cos 45 =mg but it also states that mg cos 45 is not equal to T." effectively you are saying that $T cos 45 =mg$ is not the same as "$mg cos 45 =T$". Obviously they are not the same, that's why I don't understand your question. In this case, it would be better if you took a photo of the book like you did in your last question. – NMech May 21 at 17:27
• @NMech I am happy to learn and correct my mistakes. I thank you for always guiding me in my answers. I will put the textbook name and the whole page where I have confusion(I will mark it specifically ). – Srijan M.T May 21 at 17:44
• Edit your question title - it means nothing - "why do we" what? – Solar Mike May 22 at 2:09

1. Isn't this a math problem that can be arranged as,

$$T cos 45 = mg ---> \frac{T cos 45}{cos 45} = \frac{mg}{cos 45} ---> T = mg/cos 45$$

1. There are two accelerations - centripetal (towards the center) and tangential (along the circumference).

The planes and axes of the forces

• No. I said mg * cos 45. Not divided by cos 45. – Srijan M.T May 21 at 17:04
• The simple proof is through a numerical example. Say T = 10, mg = 10*0.707 =7.07. Let's assume you are correct, that T = mg*cos 45 = 7.07*0.707 = 10? Now you can answer your own question "My Q :1 is that why mg cos 45 is not equal to tension." Can't you? – r13 May 21 at 17:38
• Ohh. Do you mean to say that there are 3 axis. Y axis is mg. Z axis is T sin 45 and mg sin 45 is X axis. Therefore , mg cos 45 - T must have some acceleration? – Srijan M.T May 21 at 17:55
• Very clear drawing you have made sir. – Srijan M.T May 21 at 17:55
• The rope has a tendency to accelerate outward (T = ma), but the tendency is restricted by Fc that pulling it back to stay on the course of rotation, and the fact that the forces in the direction of gravity are balanced. The simple experiment is for you to hold a rope with a stone wrapped at the end, and waving your arm to make a circular motion. You'll feel the tension pulling your hand, while the stone maintains on the circular path., and notice that the faster the rotation, the larger the force. – r13 May 21 at 18:22

The following is a side and a top view of the problem (hopefully you understand the sketch).

• the green is the pole and wire
• the red is the circular motion trajectory
• the black are the forces.
• with light blue are the accelerations. Please note that $$a_t=0$$, and its only for illustration purposes in the image. As a consequence there is no force component on the tangential direction. This is also supported by the question which is about uniform circular motion.

If you analyze the forces on the side view you get:

From the equilibrium on the vertical axis Z (there is no motion on the z axis , you obtain:

$$\sum F_z = m\cdot a_z$$ $$T \cdot \cos\theta - mg = m\cdot 0$$ $$T \cdot \cos\theta = mg \tag{eq.1}$$

From the (dynamic) equilibrium on the horizontal you obtain that:

$$\sum F_n = m\cdot a_n$$

Where $$a_n$$ is the centrifugal and its equal to $$\frac{v^2}{r}$$ or $$\omega^2 r$$. Therefore:

$$T\cdot \sin\theta = m\cdot a_n$$ $$T\cdot \sin\theta = m\cdot \frac{v^2}{r}$$ $$T\cdot \sin\theta = m\cdot \omega^2r \tag{eq.2}$$

If you put everything together:

$$(eq.2) \rightarrow T = \frac{1}{\sin\theta } m\cdot \omega^2r \tag{eq.3}$$ then susbstitutign the above t eq1. $$(eq.1) \rightarrow \frac{1}{\sin\theta } m\cdot \omega^2r \cdot \cos\theta = mg \tag{eq.4}$$

$$(eq.1) \rightarrow \frac{1}{\sin\theta } m\cdot \omega^2r \cdot \cos\theta = mg \Rightarrow \tag{eq.4}$$ $$\omega^2 = \frac{\sin\theta }{\cos\theta}\frac{1 }{r} \cdot g$$ $$\omega = \sqrt{\frac{\sin\theta }{\cos\theta}\frac{1 }{r} \cdot g }$$ $$\omega = \sqrt{\frac{g }{r} \tan(\theta) } \tag{eq.5}$$

Therefore substituting eq.5, to eq.3:

$$T = \frac{1}{\sin\theta } m\cdot g \tan(\theta)$$ $$T = \frac{1}{\cos\theta } m\cdot g$$ $$T = \sqrt{2}\cdot mg$$

Also as Q2: can we say there is $$a_t$$ is along z axis for this question ?

if by Z you mean the vertical direction then NO. the $$a_t$$ will be on the horizontal plane and it will always be tangent to the red circle in the image.

Very short answer: The tension in the cable must be greater than the weight since it's supporting both the weight and the centripetal force. Cosines are always less than 1.