Turning effect of forces - moments

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INeedHelpPls
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Homework Statement


A uniform board is hinged at A and supported by a vertical rope at P, 6.0m from A. A man of weight 700N stands at the end of the board at point B, which is 15.0m from hinge A. If weight of board is 50N,calculate the tension T in the rope.

Homework Equations


By principle of moments and taking moments about the hinge A, sum of ACW moments= sum of CW moments

The Attempt at a Solution


By principle of moments and taking moments about hinge A,
sum of ACW moments = sum of CW moments
700 x 15 = T x 6
T= 1750N

however, final answer is 1812.5N, can anyone help please? Thanks in advance!
 
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INeedHelpPls said:

The Attempt at a Solution


By principle of moments and taking moments about hinge A,
sum of ACW moments = sum of CW moments
700 x 15 = T x 6
T= 1750N

however, final answer is 1812.5N, can anyone help please? Thanks in advance!


You forgot the weight of the board.

ehild
 
i could not understand what you meant by ACW and CW but this question is not that hard, just take all moments about A, it must be equal to 0 since there is a static equilibrium.
taking upwards +, downwards -
6T-50.7,5-700.15=0
T=(10875)/6=1812.5N
 
Sorry ehild, while i was typing there was no answer
 
okay thanks a lot.. but i don't understand sigmaro's working :(
ACW= anti-clockwise and CW =clockwise.. something like taking upwards as + and downwards as -
 
it is nothing but rearranging
ACW=CW as ACW-CW=0 or CW-ACW=0
later on you will see that moment is a vector in fact and you will get used to this notation
 
what does the comma mean?
 
i used "." instead of "*", and so i used "," instead of "."
7,5=15/2