Physics Help Forum Torque and forces on a rigid triangle body

 Feb 28th 2017, 03:28 PM #1 Junior Member   Join Date: Feb 2017 Posts: 11 Torque and forces on a rigid triangle body Hello everyone, I am not a physicist or student so hoping someone here can help me verify my understanding of calculating torque and forces on a rigid body for a personal project I'm working on. I have the following rigid triangle, made of steel. It has 3 forces working on it. FL is a rod connected to the left triangle point to hold it down. FM is a force of 1000N pushing up on the bottom point. FR is connected to another rod which gets pushed up. It's a lever multiplying the distance FR is pushed up relative to the movement of FM. I'll post my math here in hopes someone can confirm I'm doing it right or tell me why I'm not. My ultimate goal is to find FR for the given 1000N force at FM. I'm not sure if the body of the triangle has an affect on the math... like the fact FL and FR are connected. First I use torques to get a relationship between FL and FR. FL(sin80)(15) = FR(sin70)(30) FL(14.77) = FR(28.19) FL = 1.91(FR) Now I use force equilibrium to find FR. FM = FL(cos35) + FR(cos55) FM = 1.91(FR)(cos35) + (FR)(cos55) FM = 1.56(FR) + (.57)(FR) FM = 2.13(FR) FR = 1000/2.13 = 469.48 FL = 1000 - 469.48 = 530.52 So, by exerting 1000N to the bottom of the triangle, 530.52N is applied down on the left point of the triangle (in direction of FL), holding that point in place (it pivots in reality) and the right point is pushed up in the direction of FR 469.48N. Does that look right? Attached Thumbnails
 Feb 28th 2017, 05:15 PM #2 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,130 It appears that the triangular body is supposed to be in equilibrium (not moving), correct? OK, you started with: "First I use torques to get a relationship between FL and FR. FL(sin80)(15) = FR(sin70)(30)" However, the sum of torques about any point = 0, so this should be: $\displaystyle F_L \sin(80)(15) + F_R \sin(70)(30) = 0.$ Note that either FL or FR will be negative, meaning one of these forces is pointing in the wrong direction on your diagram. Next you wrote: "Now I use force equilibrium to find FR. FM = FL(cos35) + FR(cos55)" Assuming you are analyzing the vertical components of force, you need to determine the x- and y-components of the forces FL and FR. First you have to determine the angles of FL and FR relative to the horizontal. From simple geometry the force FL is acting at 75 degrees from the horizontal, and FR is acting at 45 degrees from the horizontal, so: $\displaystyle \Sigma F_y = 0 = F_M - F_L \sin(75) + F_R \sin(45)$ and $\displaystyle \Sigma F_y = 0 = F_L \cos(75) + F_R \cos (45)$ So now you have three equations in three unknowns, and you can solve for all forces. Try this again with these corrections, and let us know what you get. topsquark likes this.
 Feb 28th 2017, 09:39 PM #3 Junior Member   Join Date: Feb 2017 Posts: 11 Yeah, I see I made at least a couple mistakes there. 1. First of all, FR should be pointing down for this free body diagram, like this: 2. I am working with vertical forces in the second step but with FL I accidentally used the angle of the moment line (which I used in the torque equation) from the vertical line. Too many lines on my paper. So, the torque equation is correct in my first post for the updated diagram of this post. After the torque equilibrium equation I have: FL = 1.91FR Then I use equilibrium of forces, working with vertical forces. FM up the other two down. 0 = FM - FLV - FRV FM = FLV + FRV FM = FL(cos25) + FR(cos55) FM = 1.91(FR)(cos25) + (FR)(cos55) FM = 1.73(FR) + (.57)(FR) FM = 2.30(FR) FR = 1000/2.30 = 434.78 I can't subtract FR from FM to get FL like I did at first because FM is a vertical force and the other 2 aren't. But I know the relationship of FL to FR. FL = 1.91FR FL = 1.91(434.78) = 830.43 How's that? I know you mentioned using the sums of verticals and sums of horizontals. Is that really necessary in this case when I already know the angle of both FL and FR relative to the vertical and torque equilibrium gave me the relationship between the two? Seems the sums would only be useful when I don't know the angle so I could find the vector using sqroot of FLH2 + FLV2. (those 2s are squares) If I need to use the vertical and horizontal, please help me understand why this way doesn't work. Thanks. :-) Attached Thumbnails
 Mar 1st 2017, 11:22 AM #4 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,130 Interesting - I did it using sum of forces in the x-direction and sum of forces in the y-direction = 0, and got different answers (I get FL = 832 and FR=429). Then I realized why: your math is correct (as is mine), but the problem is that the system is NOT in equilibrium! You cannot satisfy the three conditions of sum of forces in both x- and y-directions and sum of torque all being equal to zero with only two variables to play with (FL and FR). Go ahead and calculate the sum of forces in the x-direction using your figures for FL and FR, and you'll find it's not equal to zero. In my case I end up with sim of forces in both e directions = 0, but the torque is not equal to zero. Unless you add a third variable, such as the ability to adjust the angle of one of the forces, the system will not be in equilibrium.
 Mar 1st 2017, 11:47 AM #5 Junior Member   Join Date: Feb 2017 Posts: 11 "Unless you add a third variable, such as the ability to adjust the angle of one of the forces, the system will not be in equilibrium." So, in reality the triangle pivots on all 3 corners. M is pushed up and the triangle rotates counter clockwise around L. How does this fact change the calculations for the forces at a precise moment in time like we are doing here?
Mar 1st 2017, 12:50 PM   #6
Physics Team

Join Date: Jun 2010
Location: Naperville, IL USA
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 Originally Posted by builder89 So, in reality the triangle pivots on all 3 corners. M is pushed up and the triangle rotates counter clockwise around L.
It depends on the values you select for FR and FL. You can make it move in translation without rotating, or rotate in position.

 Originally Posted by builder89 How does this fact change the calculations for the forces at a precise moment in time like we are doing here?
If the triangle rotates then you need to define what happens to FL and FR - do they rotate as well (i.e. acting at a constant angle relative to the body)? Or do they remain operating in a constant direction (i.e acting at changing angles relative to the body)?

 Mar 1st 2017, 01:25 PM #7 Junior Member   Join Date: Feb 2017 Posts: 11 The left point, L is connected to a rod along FL. That rod has a pivot on the bottom end of it. Effectively the rod tethers the triangle down as the bottom point M is pushed up. The whole thing is a multiplier lever increasing the distance traveled at R compared to what M could travel on it's own in some given movement. M is attached to a lever with a pivot point somewhere to the left so as M is pushed up it will move along some radius but the distance it moves to the left is small relative to everything else. So, FL and FR actually will change angles as well as changing the angle they meet the triangle at because the top of FL is moving slightly left while the triangle rotates counter clockwise as FM moves up and FR move up and to the left. But for starters I wanted to simplify the situation and pick a moment in time I could do the math. Then I figure I can tackle the difference of what changes when it's moving.
 Mar 1st 2017, 05:54 PM #8 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,130 To properly analyze this assembly you would need to include the linkages that connect the triangle to the pivot point at the end of bar FL. Is FR also connected to a pivot point? What cause the force at M to be 1000N? The angles will definitely change as the triangle pivots.
 Mar 1st 2017, 08:41 PM #9 Junior Member   Join Date: Feb 2017 Posts: 11 It's the linkage in a bike suspension. I've attached a picture below to show the linkage in the context of the bike. I just pasted the original free body diagram in so the force arrows are in context of the free body diagram still (which I thought would be enough). FA is the reactive force of the ground pushing up on the rear tire. The bike IS in equilibrium as it sits still. We ignore the amount of weight resting on the front tire. The force pushing up on the rear tire (doesn't matter what it is) results in a vertical force FM on the linkage triangle. As the weight is put on the bike or it endures a fall, the increased FA force would push the right end of the swing arm up, pushing the bottom of the triangle up, which multiplies the movement at right point of triangle and FR compresses the shock. Again, we are analyzing the case of the bike standing still. The weight of the bike makes the shock compress according to the force of FR into it. So, I'm thinking I should be able to calculate FR at this moment in time, with the pieces at these angles NOW given FM as the vertical component of the force at that point. I realize it all changes as FA increases and the swing arm rotates but those calculations should be the same method, just different numbers (I think). Attached Thumbnails
 Mar 2nd 2017, 05:49 AM #10 Physics Team     Join Date: Jun 2010 Location: Naperville, IL USA Posts: 2,130 This helps. I assume the bottom corner of the triangle is pinned (not welded) to the swing arm - correct? One thing that was missing from your free body diagram is the horizontal force acting at this pivot point. Also, how is the left corner of the triangle connected to the bike frame? There seems to be a linkage that connects to the bottom of the bike frame - is the connection at the bottom end of that linkage pinned, or welded? Also, to be clear - is the only connection to the upper right corner of the triangle the shock tube (with force F_R)? In other words there is no connection from that point to the rear wheel hub, correct? Last edited by ChipB; Mar 2nd 2017 at 06:19 AM.

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