Crate lowered by two ropes and deceleration at each point of contact with rope

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SUMMARY

The discussion focuses on the analysis of a crate being lowered by two ropes, emphasizing the importance of understanding angular acceleration and the role of forces in rigid body motion. Participants clarify that only vertical forces induce torque, and thus, only perpendicular distances to these forces should be considered for calculating angular acceleration. The confusion arises from the misinterpretation of the position vector from G to A, where the Y component is deemed irrelevant in this scenario. The conclusion is that when analyzing parallel accelerations, one should focus solely on the perpendicular distances between the points of interest.

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Pipsqueakalchemist
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Homework Statement
Question and attempts below
Relevant Equations
Newton’s law
Relative acceleration
For this question, I don’t understand the steps for the relative motion part. When I try it I get angular acceleration to be zero which is obviously wrong. The solution doesn’t consider the Y distance between G and A but I don’t understand why. In the relative motion equation it’s suppose to be the position vector from G to A but if you consider that like I did it’s wrong. This is how I’ve been solving the problems so far and in rigid body kinematic so I’m very confused right now
 

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Last edited:
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Pipsqueakalchemist said:
Homework Statement:: Question and attempts below
... I’m very confused right now
Sometimes it helps to type out the problem statement :rolleyes:
When I try it I get angular velocity to be zero which is obviously wrong
This isn't about angular velocity, but about angular acceleration

And I agree with
PeroK said:
PS You should try to learn latex
Micromass said:
##\ ##
 
Last edited:
Sorry I meant angular acceleration
 
So did I do something wrong and I’m not suppose to consider the Y component of the position vector from G to A?
 
Anyone?
 
Pipsqueakalchemist said:
... The solution doesn’t consider the Y distance between G and A but I don’t understand why.
The only forces inducing a moment or torque that induces the angular acceleration are all vertical.
The relative distances to be considered for calculating that torque should be all perpendicular to those forces; therefore, horizontal distances between fulcrums and projection of vertical forces are the only significant in this problem.

WrenchTorque2.gif
 
Lnewqban said:
The only forces inducing a moment or torque that induces the angular acceleration are all vertical.
The relative distances to be considered for calculating that torque should be all perpendicular to those forces; therefore, horizontal distances between fulcrums and projection of vertical forces are the only significant in this problem.

WrenchTorque2.gif
But I’m using relative acceleration so there’s no moments or torques in the equation.
 
Is it safe to say that when using two point on an rigid body whose velocity are parallel to each other that we only consider perpendicular distance? Bc I don’t think I’ve had an problem where the two points being parallel to each other so I always consider both horizontal and vertical distances.
 
Pipsqueakalchemist said:
But I’m using relative acceleration so there’s no moments or torques in the equation.
What is the reason for the angular acceleration about G then?
TA>TB because the brake of crane A is applied more strongly than the brake of crane B.

Wherever there is a mass under acceleration (positive or negative) there is a force.
There is no force without direction and intensity.

In this case, the upward braking forces applied at A and B are different in magnitude, so one side of the crate vertically deccelerates more than the other and its center of mass vertically deccelerates at a rate between those two, and at the same time it angularly accelerates because the unbalanced pair of vertical braking forces applied 3.3 feet on its left and 3.3 feet on its right side.
Hence, the member 3.3xalpha in the solution equations (3) and (4) of the book.
 
  • #10
Lnewqban said:
What is the reason for the angular acceleration about G then?
TA>TB because the brake of crane A is applied more strongly than the brake of crane B.

Wherever there is a mass under acceleration (positive or negative) there is a force.
There is no force without direction and intensity.

In this case, the upward braking forces applied at A and B are different in magnitude, so one side of the crate vertically deccelerates more than the other and its center of mass vertically deccelerates at a rate between those two, and at the same time it angularly accelerates because the unbalanced pair of vertical braking forces applied 3.3 feet on its left and 3.3 feet on its right side.
Hence, the member 3.3xalpha in the solution equations (3) and (4) of the book.
So is it the case that since all the acceleration are parallel I can just focus on the perpendicular distances. Bc as I said above I’ve always considered both the horizontal and vertical distances when considering the position vector (r_AB) and I’ve noticed in these scenarios A and B had acceleration, velocity that weren’t parallel
 

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