# Dynamics exam prep -- forces explanation in FBD please

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I am doing a review for a Dynamics exam, but I am having trouble with this one question (below).

What I have done is make a free body diagram that includes the tension of the cable and the weight of the rider/swing. So according to my calculations for part i:

T= mg cos (30) = 680N

The tangential acceleration is 0

Which according to the provided solutions is incorrect. Am I missing a force?

I figured the tension is integral to finding the velocity of the rider. So I want to get my tension right before I proceed with finding the velocity, though I am also having trouble figuring out how to relate tension and velocity of the acceleration of the rider is 0

I would appreciate an explanation so I can understand. Related Introductory Physics Homework Help News on Phys.org
berkeman
Mentor
I am doing a review for a Dynamics exam, but I am having trouble with this one question (below).

What I have done is make a free body diagram that includes the tension of the cable and the weight of the rider/swing. So according to my calculations for part i:

T= mg cos (30) = 680N

The tangential acceleration is 0

Which according to the provided solutions is incorrect. Am I missing a force?

I figured the tension is integral to finding the velocity of the rider. So I want to get my tension right before I proceed with finding the velocity, though I am also having trouble figuring out how to relate tension and velocity of the acceleration of the rider is 0

I would appreciate an explanation so I can understand.

View attachment 197584
Welcome to the PF. if the rotational speed is zero, then theta is zero, and T=mg.

As the rotational speed increases, theta increases. So if T=mg cos(theta), the tension would be decreasing with increasing theta, which clearly is not the case. What other force acts on the rider that pulls them out? How does that force add to the tension over and above mg?

haruspex