Find max distance box on hingedbeam can be before rope snaps

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Homework Help Overview

The discussion revolves around a torque problem involving a hinged beam and a rope, focusing on the maximum distance a box can be placed before the rope snaps. The participants are exploring the forces and torques acting on the system.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations for torque and net forces, questioning whether all forces are accounted for. There is an exploration of the y-component of tension and its role in calculating torque. Some participants express uncertainty about their calculations and seek clarification on the definitions of torque.

Discussion Status

Guidance has been offered regarding the focus on the y-component of tension and the definition of torque. Participants are actively engaging with the problem, sharing their calculations and questioning their assumptions, but no consensus has been reached on the final outcome.

Contextual Notes

Participants mention that this is their first torque problem, indicating a potential lack of familiarity with the concepts involved. There is also a reference to the angle between the rope and the bar, which may influence the calculations.

isukatphysics69
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Homework Statement


iamscrewed.PNG


Homework Equations


torque = force * distance

The Attempt at a Solution



Torque forces :
TForceTensionRopeY - TBOX - TBEAM = 0

Net forces Y:
∑FY = FTensionRopeY + FHingeY - WBEAM - WBOX = 0

Net forces X:
∑FX = FHingeX - FTensionRopeX = 0

This is my first torque problem so i want to make sure that i have all of my forces properly, i have been spinning wheels here for 30 minutes unable to solve. I will continue trying but want to see if i have all of the forces correctly
 

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isukatphysics69 said:

Homework Statement


View attachment 225177

Homework Equations


torque = force * distance

The Attempt at a Solution



Torque forces :
TForceTensionRopeY - TBOX - TBEAM = 0

Net forces Y:
∑FY = FTensionRopeY + FHingeY - WBEAM - WBOX = 0

Net forces X:
∑FX = FHingeX - FTensionRopeX = 0

This is my first torque problem so i want to make sure that i have all of my forces properly, i have been spinning wheels here for 30 minutes unable to solve. I will continue trying but want to see if i have all of the forces correctly
You need the torque equation only. Write the torques with respect to the hinge. Take the tension equal to the maximum value.
 
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ehild said:
You need the torque equation only. Write the torques with respect to the hinge. Take the tension equal to the maximum value.

I believe i need the y component only of the tension tho

525 = 1.325*182 +x*225
525 - 241.15 = x*225
283.85/225 = 1.26m = x was incorrect

That is why i had the other equations so that i can find that y component of the tension​

i believe i can actually just find that y component easily 1 sec
 
isukatphysics69 said:
I believe i need the y component only of the tension tho

525 = 1.325*182 +x*225
525 - 241.15 = x*225
283.85/225 = 1.26m = x was incorrect

That is why i had the other equations so that i can find that y component of the tension​

i believe i can actually just find that y component easily 1 sec
How is the torque defined? What is the torque of the tension?
You have the angle between the rope and the bar (30 °).
 
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ehild said:
How is the torque defined?
You have the angle between the rope and the bar (30 °).
I have defined the torque as negative downwards and positive upwards
torque is force*distance

So i see what youre saying here i think, 525sin(30)*2.65
this would be the maximum force in the y direction multiplied by the distance. This is the torque of the rope
 
isukatphysics69 said:
I have defined the torque as negative downwards and positive upwards
torque is force*distance

So i see what youre saying here i think, 525sin(30)*2.65
this would be the maximum force in the y direction multiplied by the distance. This is the torque of the rope
Yes. So what did you get for x?
 
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ehild said:
Yes. So what did you get for x?
2.019 meters thank you!
 
isukatphysics69 said:
2.019 meters thank you!
You are welcome. :oldsmile:
 

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