What is the Maximum Weight of Mass C on a Tipping Barrel?

In summary, the conversation discusses the tipping of a barrel when a force is applied to it. The calculations involve using static friction to determine the maximum tension and weight of mass C, and then using kinetic friction to calculate the maximum weight of C when the barrel is sliding. The conversation also mentions the importance of choosing a helpful reference axis for calculations.
  • #1
goonking
434
3

Homework Statement


upload_2016-7-23_22-28-18.png


Homework Equations

The Attempt at a Solution


When the barrel starts to tip, the normal at point A should be 0 Newtons, and then all weight would be on point B.
I'm guessing if the barrel would tip, it wouldn't be sliding across the floor so kinetic friction is not used, instead we should use static friction.

I take the moment about G:

[(0.45m)(Ffriction)] + [(0.1m)(Ftension)] - [(0.25m)(90kg ⋅ 9.8 m/s2)] = 0

Friction Force = Fnormal ⋅ μs = 352.8 N

Solving for tension of string = 617 N

Therefore the max weight of mass C is :

T = mg
T/g = m
617N / 9.8 m/s2 = 63 kg

Is this correct?
 
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  • #2
You can ease the calculations by choosing a different axis about which to compute moments. For instance, point B might yield a simpler equation. But no matter. You've solved the more difficult equation for moments about G. Those calculations look correct. But there is a problem...
goonking said:
I'm guessing if the barrel would tip, it wouldn't be sliding across the floor so kinetic friction is not used, instead we should use static friction.

You've guessed that the barrel will tip (if it tips) without sliding. You've computed a maximum tension based on that assumption. Knowing that tension, you are now in a position to test whether that guess was correct.

So... If the barrel is subject to a rightward force of 617 Newtons (per your calculation), will it slip?
 
  • #3
jbriggs444 said:
You can ease the calculations by choosing a different axis about which to compute moments. For instance, point B might yield a simpler equation. But no matter. You've solved the more difficult equation for moments about G. Those calculations look correct. But there is a problem...You've guessed that the barrel will tip (if it tips) without sliding. You've computed a maximum tension based on that assumption. Knowing that tension, you are now in a position to test whether that guess was correct.

So... If the barrel is subject to a rightward force of 617 Newtons (per your calculation), will it slip?
yes, it will slide because I calculated static friction to be only 352 N.
 
  • #4
goonking said:
yes, it will slide because I calculated static friction to be only 352 N.
Good. Now what is the new condition for tipping or not?

[Note that your choice for the reference axis about which to compute moments has just become helpful -- we need not worry about any angular momentum associated with the sliding motion of the cylinder. Congratulations on the choice]
 
  • #5
jbriggs444 said:
Good. Now what is the new condition for tipping or not?

[Note that your choice for the reference axis about which to compute moments has just become helpful -- we need not worry about any angular momentum associated with the sliding motion of the cylinder. Congratulations on the choice]
there has to be a force that causes rotation of the body about an axis, but I'm not sure how to find the magnitude of force that will stop the sliding and cause a tipping.
 
  • #6
goonking said:
there has to be a force that causes rotation of the body about an axis, but I'm not sure how to find the magnitude of force that will stop the sliding and cause a tipping.
There is nothing that precludes tipping while sliding.
 
  • #7
jbriggs444 said:
There is nothing that precludes tipping while sliding.
Then I believe the acceleration caused by weight C has to be much greater than the deceleration caused by kinetic friction to cause a tipping while sliding?
 
  • #8
goonking said:
Then I believe the acceleration caused by weight C has to be much greater than the deceleration caused by kinetic friction to cause a tipping while sliding?
Yes. That is likely true. But the value of the acceleration does not enter into the calculations. Putting the reference axis at the center of mass of the sliding object makes that acceleration irrelevant to the question of whether the object tips or does not. Hence my earlier congratulations on having chosen a helpful reference axis.
 
  • #9
jbriggs444 said:
Yes. That is likely true. But the value of the acceleration does not enter into the calculations. Putting the reference axis at the center of mass of the sliding object makes that acceleration irrelevant to the question of whether the object tips or does not. Hence my earlier congratulations on having chosen a helpful reference axis.
since the object is now sliding, kinetic friction is used:

Fkinetic = 308.7 N

Moment about G :

-(308.7N)(0.45m) + (882N)(0.25) - (0.1) T = 0

T = 815.85 N

therefore the max weight of C is 83.25 kg

(I believe the extra 20 kg difference from my first attempt's calculation is enough to cause a tipping while sliding)
 
  • #10
Without carefully checking the arithmetic, I believe that the result is correct.
 
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  • #11
jbriggs444 said:
Without carefully checking the arithmetic, I believe that the result is correct.
The arithmetic is right. Could have been simplified by leaving g as a symbol which cancels out.
Quibble: the mass of C is 83.25kg, not the weight.
 
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What is Kinematics of Rigid Bodies?

Kinematics of Rigid Bodies is a branch of physics that studies the motion of objects that are not deformable, meaning their shape and size do not change during motion.

What is the difference between Kinematics of Rigid Bodies and Kinematics of Particles?

The main difference between Kinematics of Rigid Bodies and Kinematics of Particles is that Rigid Body Kinematics deals with objects that are not deformable and have multiple points of motion, while Particle Kinematics deals with objects that are point-like and have only one point of motion.

What are the three types of motion in Kinematics of Rigid Bodies?

The three types of motion in Kinematics of Rigid Bodies are translational motion, rotational motion, and general motion. Translational motion is when an object moves in a straight line, rotational motion is when an object rotates around a fixed axis, and general motion is a combination of both translational and rotational motion.

What is the difference between rotational and translational motion?

The main difference between rotational and translational motion is the type of movement. Rotational motion involves an object rotating around a fixed axis, while translational motion involves an object moving in a straight line without any rotation.

What are the equations used to describe rotational motion?

The equations used to describe rotational motion include angular displacement (θ), angular velocity (ω), and angular acceleration (α). These are related to linear displacement (s), linear velocity (v), and linear acceleration (a) through the equations θ=s/r, ω=v/r, and α=a/r, where r is the radius of the object's motion.

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