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Mechanics statics problem- help

  • Thread starter kyva1929
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Mechanics statics problem-- help

http://img686.imageshack.us/img686/8831/45170185.jpg [Broken]

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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tiny-tim
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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.​
 
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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!
 
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tiny-tim
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Hi kyva1929! :smile:
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:
 
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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
 
  • #6
tiny-tim
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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 …
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:
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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