Advanced Mechanics Advanced Mechanics Physics Help Forum  2Likes
Mar 2nd 2017, 11:34 AM

#11  Physics Team
Join Date: Jun 2010 Location: Naperville, IL USA
Posts: 2,179

I went ahead with an assumption that the connection at teh lower left is pinned at both ends (hence only carries force in line with the linkage). Adding in the horizontal force component at M, per the attached diagram, yield the following forces given FMy = 1000 N:
FL = 838.6 N
FR = 434.2 N
FMx = 5.5N
This satisfies all three conditions: Sum of forces in xdirection = 0, sum of forces in ydirection=0, and sum of torques = 0.

 
Mar 2nd 2017, 11:01 PM

#12  Junior Member
Join Date: Feb 2017
Posts: 11

Great. :)
Sum Fx = FRx  FLx
FMy = 1000N, but that's a vertical force.
Where did FMx come from?

 
Mar 3rd 2017, 05:27 AM

#13  Physics Team
Join Date: Jun 2010 Location: Naperville, IL USA
Posts: 2,179

FMx is the horizontal force acting on the bottom of the triangle bu the pin connection at that point. The pin connection between the triangle and the swing arm can act in any direction, not just up. It comes about because the angles of the other linkages force it to occur. As noted previously, without this the triangle is not in equilibrium. Think of it this way: if you replaced the pin connection with a sliding mechanism, so that the Fm point could slide left or right on the swing arm, it would, even though the applied force from the swing arm is in the vertical direction.

 
Mar 3rd 2017, 08:57 AM

#14  Junior Member
Join Date: Feb 2017
Posts: 11

OK, that makes sense. Thanks for your help BTW. :) Now to calculate FMx.
Using the sum of the x forces:
Fx = FLcos65 + FMx  FRcos35
FMx = 1.91FRcos65  FRcos35
FMx = .8072FR  .8192FR = 5.21
I assume the difference between my 5.21N and your 5.5N is just rounding difference in calculations.
My FL in post #3 is 830.43 and yours in post #11 is 838.6. Is this still just rounding differences in our math or something more significant?

 
Mar 3rd 2017, 10:21 AM

#15  Physics Team
Join Date: Jun 2010 Location: Naperville, IL USA
Posts: 2,179

Originally Posted by builder89 My FL in post #3 is 830.43 and yours in post #11 is 838.6. Is this still just rounding differences in our math or something more significant? 
Glad to be of help. Your post 3 failed to take into account sum of forces in the xdirection = 0. So it's not just a rounding issue.

 
Mar 3rd 2017, 12:00 PM

#16  Junior Member
Join Date: Feb 2017
Posts: 11

Can you show me where the mistake was in my math? I guess I thought I didn't need to do sum of x forces here to get FR since I was doing sum of torques first.
In post #1 I found relationship between FL and FR via sum of torques around the bottom triangle point O. Did I miss a torque here?
FL(sin80)(15) = FR(sin70)(30)
FL(14.77) = FR(28.19)
FL = 1.91(FR)
In post #3 I corrected some things and found FR. (I changed FM to FMV below cause that's what I meant originally)
0 = FMV  FLV  FRV
FMV = FLV + FRV
FMV = FL(cos25) + FR(cos55)
FMV= 1.91(FR)(cos25) + (FR)(cos55)
FMV= 1.73(FR) + (.57)(FR)
FMV = 2.30(FR)
FR = 1000/2.30 = 434.78N

 
Mar 3rd 2017, 05:37 PM

#17  Physics Team
Join Date: Jun 2010 Location: Naperville, IL USA
Posts: 2,179

The problem is that the mecahism as described in post 1 and 3 is NOT in equilibrium. You used sum of torques =0 and sum of vertical forces =0 to get your answer, but with those results the sum of forces in the xdirection is not zero. I suggest you go back and see that this is the case. The problem is that your free body diagram was incorrect  it did not treat the attachment point to the swing arm as being pinned, which would make it so that the triangle can't move left or right. A pin can apply both a horizontal and vertical force to the triangle, even though the 1000N force is purely vertical.
Perhaps a different example would help make clear why the triangle cannot be in equilibrium using only the forces of post number 3. Suppose you had a square with a force acting upward on the lower right corner of 1000N. And suppose there are two countering forces  one we'll call A is directed vertically downward at the lower left corner, and a second we'll call B is directed horizontally to the right at the upper right corner. What would be the magnitudes of A and B? To solve this you might start as you did with torques about the point where the 1000 N force is applied, and you would find that Forces A and B are equal in magnitude. Then use sum of forces in the ydirection =0 and you'll see that force A is equal to 1000N, in the downward direction. This means that force B is also 1000 N, toward the right. The problem now is that while torque =0 and forces vertically =0, the sum of forces horizontally is not 0  it's 1000N. There is no way for this square to be in equilibrium given only forces A and B. We have to add a third force into the mix in order to make it work (just as there was that third force acting on your triangle).
Last edited by ChipB; Mar 3rd 2017 at 05:40 PM.

 
Mar 4th 2017, 10:40 AM

#18  Junior Member
Join Date: Feb 2017
Posts: 11

I understand what you're saying about the 3rd force. I can visualize it must be there. That means FMy != FM. So, as you stated, my free body diagram was wrong. Actually FM is in a direction somewhere up and left on the free body diagram. I've updated it on this post.
Now I'm not sure where to start with the math.
It seems I can NOT start with the same torque equation measuring sum of torques around M. I already tried that getting FL=1.91FR.
My sum of y forces looks right to me but it yields a different result than your answer. The only way it could be wrong is if FL is NOT 1.91FR and I did my sum of torques incorrectly.
If I don't start with sum or torques like I did, I don't know where to start to get a relationship between the forces at L and R.

 
Mar 5th 2017, 03:42 AM

#19  Physics Team
Join Date: Jun 2010 Location: Naperville, IL USA
Posts: 2,179

Your torque calculation is correct. I just noticed however that I had a typo in my results in post #11; the answer shoud be:
FL = 828.6
FR = 434.2
FMy = 5.5
This yields FL/FR = 1.91, just as you have it.
So to state one more time: there are three equations and three unknowns. You cannot skip one of those equations. It seems you are now showing the force FM at an angle  assuming this is still 1000 N and assuming you know that angle, you still need to take into account the reaction force of the pin joint on that bottom corner of the triangle.

 
Mar 6th 2017, 10:37 AM

#20  Junior Member
Join Date: Feb 2017
Posts: 11

I think you meant FMx=5.5, not FMy in your last post.
But I think there is a miscommunication here.
3 equations:
1. Sum or torques
2. Sum of vertical forces
3. Sum of horizontal forces
3 unknowns:
1. FL
2. FM (though we do know the vertical component FMv is 1000N)
3. FR
We agree I did the torque equation right so:
FL = 1.91FR
In post #16, I use sum of vertical forces to find FR = 434.78N. Because only vertical forces are used in sum of forces equation, I don't care about the horizontal force from FM at that point. I KNOW FMv (not FM) is 1000N. It's a given.
So: FR = 434.78 and so FL = 830.43. If you disagree, can you please tell me why my equation in post #16 is wrong?
FR was my goal so I actually don't need to know FMx in this case but I make the attempt to be sure I am not missing something in your posts.
Now that I know FL and FR, I can use sum of horizontal forces to find FMx and I do that in post #14. I find there that FMx = 5.21. Again, please tell me if my equation there is wrong.
Finally, if I cared, I could calculate FM as:
FM = sq root ( 5.21sq + 1000sq ) = 1000.1
And even the angle of FM as:
acos(1000/1000.01) = 26°
So, using sum of torques about M and sum of vertical forces, I am able to find FR, without doing any sum of horizontal forces or finding FMx. This is because I already know FMv (the vertical component of unknown FM) as a given and I know the angles of FL and FR.
I only needed 2 equations to find 2 unknowns in getting FR.
Now I can do this for multiple FM forces into differing rotations of the triangle and linkage at different angles and for each case I would need to consider the angle of FM to point M in order to find FMv to be applied to the same steps of then finding FL and FR.

  Thread Tools   Display Modes  Linear Mode  