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

In summary: I always consider both distances.When you use relative motion, you need to consider the forces that are trying to change the velocity of the object. All of the forces that are trying to change the velocity of an object are vertical, so in this case the forces that are trying to change the velocity of the crate are the upward forces at A and the downward forces at B.The only forces inducing a moment or torque that induces the angular acceleration are all vertical.
  • #1
Pipsqueakalchemist
138
18
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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  • #2
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:
##\ ##
 
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  • #3
Sorry I meant angular acceleration
 
  • #4
So did I do something wrong and I’m not suppose to consider the Y component of the position vector from G to A?
 
  • #5
Anyone?
 
  • #6
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
 
  • #7
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.
 
  • #8
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.
 
  • #9
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
 

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

What is the purpose of using two ropes to lower a crate?

The use of two ropes in this scenario allows for a more controlled and balanced descent of the crate. It also helps to distribute the weight of the crate more evenly, reducing the strain on each individual rope.

How does deceleration occur at each point of contact with the ropes?

Deceleration occurs at each point of contact with the ropes due to the friction between the ropes and the crate. As the crate descends, the ropes create a resistance force that slows down the crate's acceleration, resulting in a gradual decrease in speed.

What factors can affect the deceleration of the crate?

The deceleration of the crate can be affected by several factors, including the weight of the crate, the tension in the ropes, and the coefficient of friction between the ropes and the crate. Other factors such as the angle of the ropes and any external forces acting on the crate can also impact the deceleration.

How can the deceleration be calculated at each point of contact?

The deceleration at each point of contact can be calculated using the formula a = F/m, where a is the deceleration, F is the frictional force between the ropes and the crate, and m is the mass of the crate. This formula takes into account the relationship between force, mass, and acceleration.

Are there any safety precautions to consider when using this method to lower a crate?

Yes, there are several safety precautions to consider when using this method. It is important to ensure that the ropes are securely attached to the crate and can handle the weight. The angle of the ropes should also be carefully calculated to avoid any sudden jerks or swings. Additionally, proper training and supervision should be provided to those handling the ropes to prevent any accidents or injuries.

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