Simillar long question which will come out during my test,

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Homework Help Overview

The discussion revolves around a mechanics problem involving forces and moments, particularly focusing on a roller against a vertical beam. Participants are attempting to analyze the forces acting on various points in a system, including reactions at supports and the calculation of magnitudes of forces.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations used to balance forces and moments, with some expressing confusion about the meaning of certain variables, such as Rc. There are inquiries about how to derive equations and the implications of having only vertical forces acting on the roller.

Discussion Status

Some participants have provided insights into their free body diagrams and force balances, while others are seeking clarification on specific calculations and assumptions. There is an ongoing exploration of the relationships between different forces and moments, with no clear consensus reached yet.

Contextual Notes

Participants mention the need for a free body diagram and the importance of understanding the geometry of the problem. There are references to specific forces and moments, as well as the challenge of interpreting the results of calculations.

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Homework Statement



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Untitled2_zps8609575f.png

Homework Equations


ƩMA = Ʃ r x F
Total Force of x,y= 0
root(rx^2+ry^2)= Magnitude

The Attempt at a Solution



I cannot find i)

my method
0=2000+3000(0.4)-Rc
-3200N=-Rc(up)

0=3200-2000-3000+Ry
Rx=1800

How to find to magnitude, when roller only got y force, i can't use root(rx^2+ry^2)
 
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cracktheegg said:
my method
0=2000+3000(0.4)-Rc
-3200N=-Rc(up)

0=3200-2000-3000+Ry
Rx=1800
I've no idea how you get these equations, or what Rc stands for. Please explain your analysis.
How to find to magnitude, when roller only got y force, i can't use root(rx^2+ry^2)
Since it's a roller against a vertical beam, I would think it can only provide an x force.
But where's the difficulty in finding the magnitude if there were only a y force? What would root(0^2+ry^2) give you?
 
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haruspex said:
I've no idea how you get these equations, or what Rc stands for. Please explain your analysis.

Since it's a roller against a vertical beam, I would think it can only provide an x force.
But where's the difficulty in finding the magnitude if there were only a y force? What would root(0^2+ry^2) give you?

They only provide 2 y force, so how i use that to find rx?
 
You haven't shown us, or at least described in words, your free body diagram. Tell us what you got. That's key to the rest of the problem.
 
My free body diagram got two reaction for at A, 1 reaction force at B at including the G2, G1 force and also one reaction force at C
 
cracktheegg said:
My free body diagram got two reaction for at A, 1 reaction force at B at including the G2, G1 force and also one reaction force at C
From the diagram, it doesn't seem that there is any reaction force at C. Next, please write for us your force balance in the vertical direction, your force balance in the horizontal direction, and your moment balance around around hinge A.
 
3(0.7)+2(0)=Rb(0.8)=0
Rb=2625


I now know how to RX RY at hinge A and answer b, but c quite difficult. i need some guidance

My guess is that I need to find the Rx Ry of the pulley and them derive the magintude of the F through Fr=root(x^2+y^2)
 
Last edited:
cracktheegg said:
3(0.7)+2(0)=Rb(0.8)=0
Rb=2625


I now know how to RX RY at hinge A and answer b, but c quite difficult. i need some guidance

My guess is that I need to find the Rx Ry of the pulley and them derive the magintude of the F through Fr=root(x^2+y^2)
You are missing the decimal point in your equation for Rb. To get Rx and Ry at A, you need to do horizontal and vertical force balances.

Part c starts out with a geometry problem. You need to find out how far to the right point C is relative to point A, and you need to find out how far point C is above point A. Can you do the geometry to find out those distances?

Chet
 
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Can check my working
taking moment at A=2000cos35 (0.4) + 3000sin83.81(1.06) / 1.6(sin64.74) = (wrong answer)

can anyone help me check

AD= 1.6
AG2=1.06
 
Last edited:
  • #10
cracktheegg said:
Can check my working
taking moment at A=2000cos35 (0.4) + 3000sin83.81(1.06) / 1.6(sin64.74) = (wrong answer)

can anyone help me check

AD= 1.6
AG2=1.06

I agree with your moment arm to G1, but, for the moment arm to G2, I get (0.8cos35+0.7sin35)=1.06

Also, you left out the moments about A resulting from the horizontal and vertical components of rope tension at C.
 
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  • #11
thanks
Chestermiller i have solve the question.
I appreciate your patient with me a lot
 

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