Static equilibrium, diving board

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SUMMARY

The discussion focuses on calculating the forces exerted on two supports of a diving board under static equilibrium conditions. A person weighing 620 N stands at the end of a 350 N diving board, supported at x = 0 m and x = 2 m. The correct calculations for the forces at the supports are FA = 1017.5 N (upward) at x = 0 m and FB = 1987.5 N (upward) at x = 2 m. The initial confusion arose from misinterpreting the question regarding the forces acting on the supports versus the forces exerted by the supports on the diving board.

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BikeSmoth
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Homework Statement


A person who weighs 620 N stands at x = 5.00 m, right on the end of a long horizontal diving board that weighs 350 N. The diving board is held up by two supports, one at its left end at x = 0, and one at the point x = 2.00 m. (a) What is the force exerted on the support at x = 0? (b) What is the force acting on the other support? (Use positive to indicate an upward force, negative for a downward force.)


Homework Equations


sum T=0, T=FR
Mgp=wieght of the person
Mgb=weight of the board
FA=force at point a - the left most support
FB=Force at point b - the rightmost support

The Attempt at a Solution


I don't know what I am doing wrong, but apparently these answers are incorrect.
(a) sum T= 0 =-Mgp*Rp+(-Mgb)*Rb+FA*Ra
FA=[Mgp*Rp+Mgb*Rb]/Ra
all about point b
FA=[620N*3m+350N*.5m]/-2m
FA=-1017.5N
(b)FB=[Mgp*Rp+Mgb*Rb]/Rb
about point a
FB=[620*5+350*2.5]/2
FB=1987.5
 
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BikeSmoth said:

Homework Statement


A person who weighs 620 N stands at x = 5.00 m, right on the end of a long horizontal diving board that weighs 350 N. The diving board is held up by two supports, one at its left end at x = 0, and one at the point x = 2.00 m. (a) What is the force exerted on the support at x = 0? (b) What is the force acting on the other support? (Use positive to indicate an upward force, negative for a downward force.)


Homework Equations


sum T=0, T=FR
Mgp=wieght of the person
Mgb=weight of the board
FA=force at point a - the left most support
FB=Force at point b - the rightmost support

The Attempt at a Solution


I don't know what I am doing wrong, but apparently these answers are incorrect.
(a) sum T= 0 =-Mgp*Rp+(-Mgb)*Rb+FA*Ra
FA=[Mgp*Rp+Mgb*Rb]/Ra
all about point b
FA=[620N*3m+350N*.5m]/-2m
FA=-1017.5N
(b)FB=[Mgp*Rp+Mgb*Rb]/Rb
about point a
FB=[620*5+350*2.5]/2
FB=1987.5
Your numbers are OK (might want to round them off), but read the question carefully. It asks for the force on the supports, not the force of the supports on the diving board (think of Newton 3).
 

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