Mechanics statics problem- help

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Discussion Overview

The discussion revolves around a mechanics statics problem involving the maximum weight that can be pushed up an incline by a tractor, considering factors such as friction, moments, and forces acting on the system. Participants explore various approaches to solve the problem and clarify their reasoning.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant attempts to use a moment equation about point A to determine the maximum weight but finds the calculated value significantly larger than the expected answer of 836 lb, leading to confusion about whether slipping or tipping occurs.
  • Another participant suggests that the friction force at point A must be considered and that the maximum engine force becomes irrelevant once the rear wheels slip.
  • A participant proposes a method to calculate the force required to move the log, using specific equations and assumptions about the direction of friction forces, but expresses discomfort with these assumptions.
  • There is a discussion about the signs in the equations and the interpretation of friction forces, with one participant correcting another's misunderstanding about the direction of friction opposing motion.
  • Clarifications are made regarding the origin of certain terms in the equations, such as the factor 0.666W, and the need to consider the free body diagram (FBD) of the tractor alone.
  • Participants reflect on their understanding of the problem, with one admitting to misinterpreting the question and recognizing the correct direction of friction forces in the context of the tractor and log movement.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to the problem, particularly regarding the assumptions about friction and the application of moment equations. There is no consensus on a single method or interpretation, and the discussion remains unresolved.

Contextual Notes

Participants highlight potential confusion regarding the direction of friction forces and the assumptions made in their calculations. There are also references to specific angles and forces that may not be fully defined, leading to uncertainty in the problem-solving process.

kyva1929
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Mechanics statics problem-- help

http://img686.imageshack.us/img686/8831/45170185.jpg

I've tried to solve for the maximum weight using moment equation about point A, and it gives way too large value of W as compared to the answer 836lb, so I assume the slipping rather than tipping would occur. But then there is no information about the force that can be developed by the engine. I'm completely lost.

Thank you!
 
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kyva1929 said:
I've tried to solve for the maximum weight using moment equation about point A, and it gives way too large value of W as compared to the answer 836lb, so I assume the slipping rather than tipping would occur. But then there is no information about the force that can be developed by the engine. I'm completely lost.

Thank you!

Hi kyva1929! :smile:

Why are you taking moments about A (the rear wheels)?

You need to take the friction force at A into account.

The maximum engine force is irrelevant … once the rear wheels slip, any extra engine force makes no difference.

Find the normal force at A, and the friction force, for a particular weight W of the log.​
 


tiny-tim said:
Hi kyva1929! :smile:

Why are you taking moments about A (the rear wheels)?

You need to take the friction force at A into account.

The maximum engine force is irrelevant … once the rear wheels slip, any extra engine force makes no difference.

Find the normal force at A, and the friction force, for a particular weight W of the log.​

I'm able to solve for the correct maximum weight using the following equation:

Force required to move the log: W*cos(10)*0.5 + W*sin(10)
taking moment about point B: -N_a*10 + 0.666W + 8000*cos(10)*(3) + 8000*sin(10)*2.5=0
summation of force about the axis parallel to the inclined: W*cos(10)*0.5 + W*sin(10)+8000*sin(10)= N_a * 0.7

The assumption in the above equation was that the friction points downward along the inclined on the log, and the friction points upward along the inclined at point A. But I am not entirely comfortable with this assumption, can you explain why it would work? (I was tempted to make this assumption, but am not comfortable with it even though I got the answer correct.)

Thank you so much!
 
Hi kyva1929! :smile:
kyva1929 said:
summation of force about the axis parallel to the inclined: W*cos(10)*0.5 + W*sin(10)+8000*sin(10)= N_a * 0.7

Yes, except one of the signs is wrong (see below).

(and "about" means rotation … say "along" or "parallel to" or "in the direction …")
taking moment about point B: -N_a*10 + 0.666W + 8000*cos(10)*(3) + 8000*sin(10)*2.5=0

Where does 0.666W come from? :confused:
Force required to move the log: W*cos(10)*0.5 + W*sin(10)

What is this equation for? :confused:
The assumption in the above equation was that the friction points downward along the inclined on the log, and the friction points upward along the inclined at point A. But I am not entirely comfortable with this assumption, can you explain why it would work? (I was tempted to make this assumption, but am not comfortable with it even though I got the answer correct.)

Friction always opposes motion. Both friction forces will be in the same direction.

Start again! :smile:​
 


tiny-tim said:
Hi kyva1929! :smile:Yes, except one of the signs is wrong (see below).

(and "about" means rotation … say "along" or "parallel to" or "in the direction …")Where does 0.666W come from? :confused:What is this equation for? :confused:Friction always opposes motion. Both friction forces will be in the same direction.

Start again! :smile:​

That 0.666W came from (0.5 cos(10) + sin(10)), the reaction force from log, and I forgot to multiply it by 1.25ft. The equations listed was based on the FBD of only the tractor.

The direction of friction is indeed confusing to me in this problem. When the problem ask to find out the maximum weight that can be pushed up, I assume the frictional force acting on the log points downward along the inclined. And to consider the frictional force acting on the rear wheels, we are considering the extreme case when the rear wheels are just about to slip, the frictional force should point upward along the inclined, or else there will be no force supporting the tractor; all forces acting on the tractor points downward along the inclined.

Using the equations I listed before (0.666*1.25 W instead of 0.666W), it gives me the answer given by the book. So was that the error of the book or I just arrive at the correct answer by accident?

Thank you so much
 
oops!

hmm, I think I've been misinterpreting the question :redface: …

the force isn't an external force, it's coming from the tractor engine trying to turn the back wheel, isn't it? :rolleyes:

so what I said was wrong …
tiny-tim said:
Friction always opposes motion. Both friction forces will be in the same direction.

Friction does oppose motion, but of course the log is moving up, while the engine is trying to move the rear wheel (well, the bit in contact with the road) down …

so you were right, the friction forces are in opposite directions! :redface:
kyva1929 said:
That 0.666W came from (0.5 cos(10) + sin(10)), the reaction force from log, and I forgot to multiply it by 1.25ft. The equations listed was based on the FBD of only the tractor.

Yes, that seems ok now! :smile:
 

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