Solving a Homework Problem with One Pivot

[Falcon99]

Homework Statement

https://cdn.discordapp.com/attachments/373928512811302912/528638926148337704/question.PNG

part a )50N
I cannot answer part b.

Homework Equations

M=FD

The Attempt at a Solution

Ok so if both pivots were in contact with the beam I would just equate them with the oppposite moments( anticlockwise ,clockwise) e.g 4
However because only 1 pivot is in contact with the beam I can't figure out how to equate it . I tried finding the resultant anticlockwise moment which was 100nm but that would me it would be turning so that's not right. This website http://wiki.math.se/wikis/2009/bridgecoursemechanics/index.php/14._Moments_and_equilibrium has been usefult but only explains it when there is 2 pivots .

 

[Doc Al]

There are two facts that you can use:
(1) The net torque about any point must be zero. (Assuming equilibrium.)
(2) The net force on the plank must be zero.

That should allow you to solve b.

 

[PhanthomJay]

The problem is not worded correctly, it should say that the board is just on the verge of lifting off the right support.
But how did you-get the answer to part a?? Please show your work.

 

[Falcon99]

The problem is not worded correctly, it should say that the board is just on the verge of lifting off the right support.
But how did you-get the answer to part a?? Please show your work.

I did 2Xw=200X0.5
w=50N

 

[Falcon99]

I did 2Xw=200X0.5
w=50N

There are two facts that you can use:
(1) The net torque about any point must be zero. (Assuming equilibrium.)
(2) The net force on the plank must be zero.

That should allow you to solve b.

There are two facts that you can use:
(1) The net torque about any point must be zero. (Assuming equilibrium.)
(2) The net force on the plank must be zero.

That should allow you to solve b.

How do I find the force though , I have equated those but just proves they're are equlibrium like you said, I can't see where I can equate for F(force)

 

[Doc Al]

How about you use fact #2 from my list? Express that mathematically.

 

[Falcon99]

How about you use fact #2 from my list? Express that mathematically.

Ok I've found the anticlockwise and clockwise moments, so net moment is 0 . But still haven't found a way to prove the upward and downward force on the plank is 0.

 

[Doc Al]

But still haven't found a way to prove the upward and downward force on the plank is 0.

That's just a condition for equilibrium, which must be assumed to solve the problem. (As @PhanthomJay points out, the problem could have been worded more clearly.)

 

[Doc Al]

(Hint: Clearly identify the forces acting on the plank. There are three: Two act down, one acts up.)

 

[haruspex]

haven't found a way to prove the upward and downward force on the plank is 0.

You do not need to prove that. As @PhanthomJay pointed out, the question is poorly worded. It should say that the other end only just starts to lift, i.e. the acceleration is negligible. This allows you to treat it as a statics problem, so the forces are necessarily in balance. You just need to express that as an equation.
Note that the torque balance equations you already used assume the system is effectively static. It cannot be solved without that assumption.

By the way, it is not strictly necessary to use the linear force balance. You can solve the problem entirely with torques, provided you use two different axes. Indeed, the linear balance is equivalent to using an axis at infinity.

 

[Falcon99]

You do not need to prove that. As @PhanthomJay pointed out, the question is poorly worded. It should say that the other end only just starts to lift, i.e. the acceleration is negligible. This allows you to treat it as a statics problem, so the forces are necessarily in balance. You just need to express that as an equation.
Note that the torque balance equations you already used assume the system is effectively static. It cannot be solved without that assumption.

By the way, it is not strictly necessary to use the linear force balance. You can solve the problem entirely with torques, provided you use two different axes. Indeed, the linear balance is equivalent to using an axis at infinity.

Yeah all makes sense now , I see what you mean I can just equate like a normal statics question. Thanks a lot for everyones help and time.