Thread: Statics Problem. Reaction forces, etc.

We have a test coming up next week and here's a practice problem I have no idea how to do. Mainly the way the pulley is holding the weight throws me off.

For these type of problems, we are going to have to solve for the reaction forces, and show them as vectors.

I drew my free body diagram.
The forces in the x and y have to be equal to 0, since this is statics.

Can someone show me how to do this problem. I really don't know where to start, and we haven't went over one like this in class. Any information you guys can provide will be extremely helpful, as I'm having a hard time with this class.

Originally Posted by Bracketology

We have a test coming up next week and here's a practice problem I have no idea how to do. Mainly the way the pulley is holding the weight throws me off.

For these type of problems, we are going to have to solve for the reaction forces, and show them as vectors.

I drew my free body diagram.
The forces in the x and y have to be equal to 0, since this is statics.

Can someone show me how to do this problem. I really don't know where to start, and we haven't went over one like this in class. Any information you guys can provide will be extremely helpful, as I'm having a hard time with this class.

List the forces, there are two reaction forces and two forces corresponding to the tension in the string that need to be accounted for, these sum to zero.

Take moments about the left hand pivot, these should also sum to zero.

Now if you knew the horizontal distance of the pulley axle from the left hand pivot you would be able to solve for all the forces.

CB

I emailed the teacher and he said to assume the middle of the pulley and the far right upward reaction force from the roller joint are in the same line. Now the problem makes a little more sense.

Thanks for the help!

So, is this anywhere near correct?

Forces in X=0=Ox (Reaction from the wall in x)+1500sin(18.435)= Ox=-474.343 Newtons
Forces in Y=0=Ox (Reaction from the wall in y)+1500cos(18.435)=Oy=1423.03 Newtons
Moment About Point O (Origin)=0=-(2.0M)(1500KG)(N*S^2/KG*M)=3000N*M

Honestly lost on this problem. We have been doing very simple problems in class, and this is way beyond what we have went over, and we have to put all that we know so far together to get this answer and I'm having trouble. Any more direction on what to do would help.

Originally Posted by Bracketology

So, is this anywhere near correct?

Forces in X=0=Ox (Reaction from the wall in x)+1500sin(18.435)= Ox=-474.343 Newtons
Forces in Y=0=Ox (Reaction from the wall in y)+1500cos(18.435)=Oy=1423.03 Newtons
Moment About Point O (Origin)=0=-(2.0M)(1500KG)(N*S^2/KG*M)=3000N*M

Honestly lost on this problem. We have been doing very simple problems in class, and this is way beyond what we have went over, and we have to put all that we know so far together to get this answer and I'm having trouble. Any more direction on what to do would help.

The only forces that are relevant are vertical, we have reaction at left hand pivot , and at right ball . We have the tension in the cable supporting the mass is , and this acts on the pully twice so we have:

(positive upwards)

Taking moments about the left hand pivot/support:



CB

Thank-you sir.