Static equilibrium, diving board

AI Thread Summary
The discussion revolves around a physics problem involving static equilibrium on a diving board supported at two points. The individual attempts to calculate the forces exerted on the supports, with initial calculations yielding incorrect results. The key equations involve summing torques and applying Newton's third law, emphasizing the need to consider the forces acting on the supports rather than the forces they exert on the board. Clarification is provided that the question asks for the forces on the supports, which requires a different approach to the calculations. Accurate understanding of the problem is crucial for obtaining the correct answers.
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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