physics

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

moneyking

Full Member
7+ Year Member
Joined
May 20, 2015
Messages
40
Reaction score
1
upload_2015-7-22_17-25-41.png


Can someone explain 31 please. Thanks
 
Since the board is in static equilibrium, there is no rotation around the point where the rope is attached. So the rotation caused by the mass (counter-clockwise) is equal magnitude and in opposite direction of the rotation caused by the weight of the board (clockwise). The weight of the board can be treated like a point for at the center of mass, 0.5 meters in, similar to this picture:
moments_wl_eg5.gif


So 0=(moment of board) - (moment of mass)
0= Fboard*(distance to T) - Fmass*(distance to T)
0=(m*10)*0.3 - 3*10*0.2
Solve for m, m=2 kg
 
"The weight of the board can be treated like a point for at the center of mass"
can you elaborate please
 
Sure, so if you have an object like this:
fig%201.jpg


Gravity acts on all masses, so at every little point along the bar, gravity is acting, shown by all of the little force arrows. To simplify this for calculations, we can represent all of these little forces as one big force at an object's center of gravity.
Like this:
centre-of-gravity.png
 
Top