Mechanical Advantage of bent lever arm

snipped-for-privacy@hotmail.com (Gord C) wrote in news: snipped-for-privacy@posting.google.com:

I would have thought school was out by now, but in any event... When you are drawing things in ASCII it is always a good idea to remind people that it is best viewed in a fixed font. Also, be careful of your terminology. MA, by which I assume you mean moment arm, is not a standard abbreviation. It is always wise to error on the side of caution when communicating by text.

I assume this drawing is telling me that the elbow of the lever is 2.5" below the pivot and the force is applied 8" from the elbow, the stop is 3" from the location of the applied force and that portion of the lever is horizontal when on the stop.

The magic of statics is that it is fundamentally based on f=ma. If, as in this case, there is no motion going on all the moments and all the forces must be balanced out to 0. I think your problem is you are being a little fuzzy or sloppy on the notion of a moment and a moment arm. A moment is always around some particular point. Your statement, "The MA of the lever appears to be 1.6..." is incomplete. You should say something like, "The moment arm around point A is 1.6"...", with that point labeled on your drawing. Having picked a point, you should sum all the moments around that point. They should sum to zero. If there is more than one unknown, you will have to pick another point or the horizontal or vertical forces to sum. Eventually, you will get to the point where you have enough equations to solve for the unknowns. In a simple problem like this you can usually get away with a single equation if you are clever at picking the point. Go back to the problem now and work on it, if you are still having difficulty, read the next paragraph.

As a starting point sum all the moments around the elbow of the linkage. Once you have done that, compare it to picking the pivot and summing all the moments around it. I have been completely unable to figure out where you got 1.6".

Charly Coughran snipped-for-privacy@REMOVE-TO-REPLY.UCSD.EDU

That's incorrect. 1.6 in. is not the right moment arm for either force.

Remember M = r X F, i.e. a moment arm is perpendicular to the force. For the 65 lb down and reaction up forces that would be horizontal distances. Sum moments about the pivot point. Figure it out...

The laws that govern the straight lever also apply to the bent lever. In this case, it is considered a second class lever where the weight (Fy) is placed between the fulcrum (o) and the force (v). In the case of the bent lever, care must be taken to determine the true length of the lever arms. In every case the true length of the arms will be the perpendicular distance between the fulcrum and the direction line of force or weight. From your sketch (I don't have Statics by Riley), I must assume that the lever is a perfect 90 degree "L" with the Force (v) and Weight/Reaction (Fy) vertical as drawn.

In this case, the sum of the moments of the forces = 8 x 65 - 5 x Fy = 0, where Fy is the weight/reaction located 3" to the left of the applied force, v. the sum of the moments of the forces = 520 - 5Fy = 0 Fy = 520 / 5 = 104

This is the same concept of a pry bar that moves a great weight located near the end that does the prying. The longer the bar the greater the load that can be moved. At one time, railroad workers would start a box car moving by placing a bar beneath one wheel and lift to get it roll, moving the boxcar.

Hope that helps, Jim Y

stand for Mechanical Advantage (1.6) so that the reaction at the stop is 65x1.6 = 104. However, in completing the question Q9.13 from Statics (Riley, 1993) my answer of 174lbs does not agree with his answer (of 181). Anyone who has this book and would like to tackle it, please have a go. Many thanks, Gord

snipped-for-privacy@hotmail.com (Gord C) wrote in news: snipped-for-privacy@posting.google.com:

Fair enough, mechanical advantage is the ratio of the moment arms so yeilds the same answer Jim, you, and I obtained. (Although I do have a minor quibble with both of you, since the problem defines the problem as

You are correct that the answer should have been a negative for a reaction. During my years of employment I can't say that I have ever seen a comparable problem showing a negative answer. What was shown on all of our calculation sheets was a notation indicating *up*,

Jim Y (retired)

Yep, R = 65lb*8in/5in = 104lb.

And it won't be the last time a text is in error.

Jim Y

The complete problem (William Riley et al, Statics, pub Wiley 1993) is reproduced on my webpage at

one pound FORCE = 4.448222 Newtons

Jim Y

The question of the lever has been settled thanks to Jim and others. As you can see from earlier postings the lever was only part of a more detailed problem which is posted at

I have appended my analysis to this webpage and got my answer of

This isn't right, because the lever in question isn't a stick

Here's the original diagram, modified to look a little more proportional: o pivot | 2.5" | | L---------------T----------v (force=65) 5" | 3" |

There can't be just a downward force there - when you pull straight down at point v, it's oging to want to rotate at point o, so point L wants to move to the left. Howcome nobody's accounted for that yet? And when is somebody going to notice the hypotenuse?

I don't know what the answer is, but it's not 5/3. (or 8/3 or 8/5 either, for that matter.)

Since my background is electronics, I started solving it with vectors, but I'm way too lazy to do all that arithmetic. :-)

But in blahblahblah, the force at v applies a torque at o, which translates into a torque x at point T.

The hypotenuse of a 2.5" x 8" right triangle is h, and the Y component of the force at V is 65. And here I don't know how to get the math right, but it's almost intuitive that it's going to pull to the left. If that's not calculable, I'll eat my hat - probably as simple as a look-up table somewhere, but I slept or drank my way through that particular class. ;-)

Good Luck! Rich

The vertical distance does not matter. The vertical distance and the vertical force have a zero cross product. See M = R X F in a statics book.

Hm, try again.

The 2.5 dimension is irrelevant. The book probably put that in there to throw you off and see if you can cut away the unnecessary data, since the force is parallel to the 2.5 dimension arm.

Sum of the moments = 0 = (-65lbs * 8ft) + (Fy-lbs * 5ft) Fy= +104 lbs

That's the force acting on the block via the lever normal to the surface of the lever, not the answer to the problem itself.

BTW, forces acting down or to the left are "generally" defined as negative, unless you're using a unique coordinate system. But so long as you keep your directions (positive/negative) consistent, it won't matter.

Dave

Right, Dave. The book I have been referring to has many questions where irrelevant data has been thrown in. I think this has tricked a few people (including me) into thinking a bent (90 deg) lever is something special. OK, this still leaves the whole friction problem to be solved. My answer is on the website at

I considered friction between the lever and block. Problem states that the coeffienent of friction (mu) is 0.2 on all surfaces.

Taking the lever and summing moments about the hinge, I get,

-65lb(8ft) + R(5ft) - mu*R(2.5ft) = 0 (counterclockwise is positive).

Solving, R = 115.55 lb. Note this force acts down on block.

Now summing forces on the block.

Along plane: 120lb + mu*R + mu*N - Wa*sin(15deg) = 0 (N is normal force on block)

Normal to plane: -R + N - Wa*cos(15deg) = 0.

Rearranging equations so Wa = Wa, I get N = 2564 lb. This yields, Wa = (N - R)/cos(15deg) = 2534.8 lb (min).

Damn, you're right... I missed that one. I was was nelgecting friction. Guess the 2.5 does make a difference in this case.

Problems like this need a good practical sense of what's going on before writing any equations of equilibrium. In this problem, the question is "what is the Wmin to prevent slipping". In this case anything less than Wmin will result in the box being pushed up the plane. Therefore the limiting frictional forces (mu*R and mu*N) act downwards along the plane; Jeff has these reversed to give Wmax, above which the box slides down. Also, (I have fuzzy thinking here), Jeff's analysis has the moment of friction acting on the lever in the wrong sense so that:- -65lb(8ft) + R(5ft) + mu*R(2.5ft) = 0 (counterclockwise is positive). which gives R=94.55lbs Gord.

A positive F1 should act up on the lever and downward on the block.

The problem states the min Wa to prevent the block from sliding Down the slope. So sliding downward is the direction of motion, which friction acts against.