# Average Tension in the string of a pendulum: Determining radial component of weight

Self-studying Classical Mechanics right now and was working through an example until I got to a point where I felt certain questions need be addressed. I will list the problem statement, the example up to the point in which I decided I need to clarify some things and then my question below.

The first part of this question series is linked here: Average Tension in pendulum string: Understanding the radial $F = ma$ equation

Problem Statement

Is the average (over time) tension in the string of a pendulum larger or smaller than $$mg$$? By how much? As usual, assume that the angular amplitude $$A$$ is small.

Relevant portion of the example

Let $$\cal{l}$$ be the length of the pendulum. Then the angle $$\theta$$ depends on time like: $$\theta(t) = A\cos(\omega t)$$ Where $$\omega = \sqrt{\frac{g}{\cal{l}}}$$

Since $$T$$ is the tension in the string the radial $$F = ma$$ equation is: $$T - mg = m\cal{l}\dot{\theta^2}$$

My Question

My question involves determining the radial component of the weight. Since we intend on utilizing the radial component of the $$F = ma$$ equation and physically, by inspection, we know that the only forces on the string in the radial direction are the tension $$T$$ and the radial component of the weight $$W_{rad} = -mg$$. $$W_{rad}$$ has been given in the example as $$W_{rad} =-mg\cos\theta$$ however when calculating this myself I ended up getting $$W_{rad} = -\frac{mg}{\cos\theta}$$ I have provided a figure below outlining my strategy: Figure $$1$$ is simply a physical drawing of the system. Figure $$2$$ is my attempt at listing the forces on a piece of the string. Figure $$3$$ I have rotated the bottom triangle to make it clearer that I was utilizing right triangle trigonometry (SOHCAHTOA) to determine the value of the hypotenuse. Where I utilized $$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{-mg}{W_{rad}}$$ and simple algebra leads me to my result. So, my question is where exactly did I go wrong? And more generally, are there any good resources for determining various components of forces? As I tend to be extraordinarily bad at determining the various components of forces given a physical system.

this is the free body diagram

take the sum of the forces toward $$~\mathbf e_T~$$ you obtain the tension force $$~T$$

$$T=m\,L\,\dot\theta^2+m\,g\cos(\theta)$$

and take the sum of the torques about the point A you obtain the equations of motion

$$m\,L^2\,\ddot\theta+m\,g\,L\sin(\theta)=0\quad\Rightarrow\\ \ddot\theta+\frac gL\sin(\theta)=0$$

As far as the weight force is concerned, mg is the magnitude of the entire weight vector, not the vertical component. IF the string were horizontal, then then the radial component of weight would be zero. W_rad is the radial component of mg: W_rad = mg*cos(𝜃), directed away from the pivot.

• How can I use right triangle trig to determine that the radial component of weight is in fact $mg\cos\theta$? In other words, where did I go wrong in determining the radial component above. I feel as though I set up and executed the trig correctly however, I clearly must not have. As well, I often have trouble with determining components of forces in general. So, this is a deep-rooted misunderstanding I have about this particular class of physical intuition. Commented Jan 21, 2023 at 5:32
• The hypotenuse of your right triangle is the total force vector. The hypotenuse (the side OPPOSITE the right angle) is vertical. The two component directions are of your right triangle would be parallel to the string and perpendicular to the string. Your right angle will be pointing rightward in your picture above. Commented Jan 23, 2023 at 23:14