Can I get help solving this bottle opener problem using torques?

[Soicowboy]

Homework Statement

Learning from home over quarantine, need help finding online learning material

Relevant Equations

Torque / rotational force?

Hi guys I'm studying at home over lock down would anybody be able to point me in the direction of some online learning material or what concept I should look up to solve this problem and find formulas . I'm stuck and can't find anything relevant online while searching.

Thanks guys any help is great I'm not really a physics guy this if for a fabrication course I'd normally get help at a night school

 

[berkeman]

Welcome to the PF.

Are you familiar with drawing free body diagrams (FBDs)? If so draw one for this problem (call the right end where the force is applied "C").

And with FBDs, are you familiar with using the sum of torques and sum of forces to figure out any accelerations, etc. for the object represented by the FBD?

I believe that for this problem you will just set the sum of the two torques B-C and B-A to zero to find the force at B...

Does that help?

 

[Soicowboy]

Thanks a lot I'll look up and learn those concepts and post my working. Appreciate your time thank you

 

[berkeman]

Thanks a lot I'll look up and learn those concepts and post my working. Appreciate your time thank you

Great, we look forward to that. Just to clarify a bit, the two torques on the bottle opener (which you will show in your FBD) are the torque from the 50N force applied at "C" about the fulcrum point at "B", and a counter torque from the force between the opener and the bottlecap at "A" about the same fulcrum point "B".

 

[Soicowboy]

 

[haruspex]

View attachment 262300

Both forces act about the same fulcrum. How far is the 50N pull from the fulcrum?

 

[Chestermiller]

. No. The pivot point is at B. So the moment arm for the 50N force is 0.085 m.
View attachment 262300

 

[Soicowboy]

0.085m

 

[Soicowboy]

Ok thank you I'll rework.

 

[Soicowboy]

CM = CCM

A x 0.025 = 50N x 0.085
A x 0.025 = 4.25NM

A= 4.25 / 0.025

= 170N?

 

[SammyS]

CM = CCM

A x 0.025 = 50N x 0.085
A x 0.025 = 4.25NM

A= 4.25 / 0.025

= 170N?

Yes.

 

[Soicowboy]

Would the same equation and using a FBD and sum of forces be applicable to this problem.

Really appreciate your guys / girls time thank you.

 

[Chestermiller]

Would the same equation and using a FBD and sum of forces be applicable to this problem.

Really appreciate your guys / girls time thank you.

Sure

 

[Soicowboy]

Sure

 

[Soicowboy]

..

 

[haruspex]

..

Close, but at one point you turned 0.05 into 0.5.

 

[Soicowboy]

Close, but at one point you turned 0.05 into 0.5.

70 / 0.05
= 1400N ?

For part b of the problem. Would you divide the first anwser to find out how much more force is required to generate another 100N.

Thank you so much!

 

[haruspex]

= 1400N ?

Yes. It's always a good idea to do a sanity check. Would you expect the force to be more or less than the applied 200N?

For part b of the problem

Part b? Of which problem?

 

[Soicowboy]

Yes. It's always a good idea to do a sanity check. Would you expect the force to be more or less than the applied 200N?

Part b? Of which problem?

I think I solved it ?

 

[Soicowboy]

Yes. It's always a good idea to do a sanity check. Would you expect the force to be more or less than the applied 200N?

Part b? Of which problem?

Haha yup that makes total sense, wasn't being logical and focusing properly.

 

[haruspex]

I think I solved it ?

Your FBD for 6a was wrong.
You need the perpendicular distance from the force to the fulcrum. For P, that means the distance from the fulcrum point to the infinite straight line which is the line of action of P. The exact point where P is applied is not relevant.

Edit: besides, the 300mm and the 50mm are not in the same direction, so it makes no sense to add them.

 

[Soicowboy]

Your FBD for 6a was wrong.
You need the perpendicular distance from the force to the fulcrum. For P, that means the distance from the fulcrum point to the infinite straight line which is the line of action of P. The exact point where P is applied is not relevant.

Edit: besides, the 300mm and the 50mm are not in the same direction, so it makes no sense to add them.

So you wouldn't include the 50mm on the line and just work off 300mm?

 

[haruspex]

So you wouldn't include the 50mm on the line and just work off 300mm?

Right.

 

[Soicowboy]

Right.

 

[haruspex]

6a is right now, but I do not understand your working in part b.
What is the 250?
How did you get the last line?
The trouble with just posting numbers is that it gives me no insight into what you are doing. Please explain your steps.