In a simple pendulum, $\alpha$ is angle between string and vertical, $T$ is tension in string and $mg$ be pull on the bob.
$T$ has 2 component $T\cos\alpha$ and $T\sin\alpha$ $mg$ has 2 components $mg\sin\alpha$ and $mg\cos\alpha$. I get that $mg\cos\alpha=T$ and $mg=T\cos\alpha$. So I got 2 values of tension at a same time $mg\cos\alpha$ and $mg/\cos\alpha$. I could not figure out the value of tension.

  • $\begingroup$ Draw a free body diagram, it will help you. $\endgroup$
    – Gert
    Commented Jun 18, 2020 at 14:40
  • $\begingroup$ Do you need guidance to your answer ? If yes , let me know or else accept an answer. Thank you $\endgroup$
    – S.M.T
    Commented May 26, 2021 at 7:00
  • $\begingroup$ Animation $\endgroup$
    – Shub
    Commented Jan 20, 2022 at 17:27

2 Answers 2


As you know, it is generally easier to resolve forces into components at right angles to each other. Obvious possibilities (in this two dimensional set up) are horizontal and vertical or along the string and a right angles to it.

Suppose that the simple pendulum has been displaced through angle $\theta$ and has just been released. In that case its velocity is zero and there will be no centripetal acceleration, that is no acceleration in the string direction.

In that case it makes sense to resolve forces along the string and at right angles to it. Along the string, because there is no acceleration, we have $$mg \cos \theta =T$$ [If the pendulum is moving, the body has a centripetal acceleration and the equation becomes $T - mg \cos \theta = \frac {mv^2}{\ell}$, but for many purposes we can neglect this centripetal term.]

At right angles to the string we have $$mg \sin \theta =ma$$ in which $a$ is the 'tangential' acceleration of the pendulum bob.


The setup you describe is 2-dimensional, and therefore the tension $\mathbf{T}$ is a vector, which has a component along the horizontal direction, $T_x$, and one along the vertical direction $T_y$, as $\mathbf{T}=(T_x,T_y)$.

Gravity acts in the vertical direction, so $m\mathbf{g}$ (which in general is also a vector) is purely along the $y$ direction, $m\mathbf{g}=(0,mg)$.

By combining forces along $x$ you get the $T_x$ component of the tension, and by combining forces along $y$ you get the $T_y$ component of the tension. The total tension is a vector with these components.


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