# Simple Statics Pulley FBD Help Please

## Homework Statement

A man having a weight of 150 lb supports himself by means of the cable and
pulley shown in the figure. If the seat has a weight of 15 lb, determine the force that must
be exerted on cable at A, and the normal force the man exerts on the seat.

Answers provided by the teacher:
TA=110 lb.
NS=40 lb.

ƩFy=0

## The Attempt at a Solution

I have attached my rude attempt at a free body diagram. Sorry, this was my only option.
Basically, I am asking someone to check my free body diagram. Any help would be greatly appreciated!

#### Attachments

• PULLEYFBD.gif
11.3 KB · Views: 945

## Answers and Replies

The forces are wrong at location D. The person sitting on the seat is pulling down at A. Therefore the tension at D is not 165 lbs.

At D, going off the FBD I attached, would I... replace the two 165 lb. forces with Tc and Tc? And Tc = 165/2? I don't think that is right, but I'm not sure what else could possibly be going on?

Edit: Are you trying to say that his pull causes a force at D as well that I have not included?

Last edited:
"Edit: Are you trying to say that his pull causes a force at D as well that I have not included?"

Yes. Because he is pulling down at A, he is exerting less normal force on the seat at E. This force is unknown. Therefore the tension at E must be less than 165 lbs.

Not sure if this is right...
I left my free body diagram for C as it is in my attachment.
I changed the free body diagram for D... I now have Ta going up, and 2Tc going down with 165/2 going down?

What is the tension at E? I am looking for an algebraic expression. You are implying that it is 165/2. That is incorrect.

Provide me with an algebraic expression for the tension at E.

Hint: You are standing on a bathroom scale. Next to the scale is a rope dangling from the ceiling. You pull on the rope. What happens to the scale reading? Relate the scale reading (tension at E) of the scale to the pulling force on the rope.

Ok algebraic expression for tension at E... Te + Ns - 150 - 15 = 0?

Edit: Not trying to be rude or pushy, thank you for your help, but, if this isn't right, could you just supply me with the correct free body diagram? I have to leave for class at 3:40 EST at the latest, I'd be very grateful for that.

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OK good.

TE = 165 + Ns
where Ns is the force that you are pulling down on pulley C at A . You have two unknowns, namely TE and Ns.

Now looking at pulley C, do you see any mechanical advantage to the arrangement? Can you algebraically relate Ns to TE thereby getting a second equation. Once you have that you can solve your problem.

is the second equation Ns = 2Te?

No time to draw but here is your solution.

At E

Te = 165 - Fa
where Te is tension at E. Fa is force person pulls down with.

At A

Fa = 2 * Te
due to mechancial advantage.

Solving: Te = 55 lbs; Fa = 2 * Te = 110 lbs.

Seat force = 150 - 110 = 40 lbs

Yes thank you so much! Just in time to get to class on time too! Thanks again

Edit: Not important for class, but how did you figure Te to be 55 lbs. Just from solving the equations?

Yes, just solved equations.