Calculating the Unbalanced Force in a Pendulum System

[krypt0nite]

I don't know how to calculate the magnitude of the unbalanced force.

A pendulum bob with a mass of 1.80 kg is displaced from its equilibrium position, A, to an unstable position, B, by means of a string which is kept horizontal. The force applied to the string is, F. The length of the pendulum is 1.25 m. The angle through which the pendulum is displaced is 30.0 degrees as measured from the vertical. Determine the following.

diagram

Draw a free body diagram for the pendulum bob when it is at an angle of 12.0 degrees as measured from the equilibrium position. Ca1culate the magnitude of the unbalanced force acting upon the bob. In what direction must the unbalanced force act?

I've calculated the speed which is 1.66m/s. And from the speed I've got the centripedal force which is 3.96N. I know the unbalanced force acts toward the center.

Ok, how do show how i calculated the unbalanced force.
I was thinking T- mg(cos something)= centripedal force but unsure of how and what to do next.

 

[Doc Al]

I've calculated the speed which is 1.66m/s. And from the speed I've got the centripedal force which is 3.96N. I know the unbalanced force acts toward the center.

Careful. While there must be an centripetal force (which you calculated), it is only a component of the unbalanced (net) force. (If the unbalanced force only acted towards the center, then the speed of the pendulum would not change--it would just keep going in a circle--but you know it speeds up as it falls!)

Ok, how do show how i calculated the unbalanced force.
I was thinking T- mg(cos something)= centripedal force but unsure of how and what to do next.

First find the tension in the string by applying Newton's 2nd Law in the centripetal direction: T - mg cos\theta = ma_c = m v^2/r. (Solve for T.)

There are two forces on the pendulum bob: the weight and the tension in the string. Add them (as vectors, of course) to find the net unbalanced force.

 

[krypt0nite]

Does Fnet=5.4N 35 degrees above horizontal sound right?

 

[Doc Al]

Sounds good to me.