A boy of mass 23 kg sits on a seesaw 1.5 m from the axis of rotation. At what
distance from the axis of rotation must his friend, whose mass is 21 kg, be
positioned to balance the seesaw?
2007-03-15 07:49:07 · 3 answers · asked by Zoolander 1 in Science & Mathematics ➔ Physics
Imagine that the axis is at the origin (0,0).
Use the equation (m1+m2)r= m1r1+m2r2 where r is the distance from the axis. Since r=0 the equation becomes
m1r1=-m2r2
(23)(-1.5)=-(21)(r2)
so r2 = 1.64 m.
2007-03-15 08:00:56 · answer #1 · answered by Froggiesmiles 3 · 0⤊ 0⤋
Formula: F1*d1=F2*d2
Also recall force =mass*accelartion(gravity)
Mass = 23kg Force = 23*9.81ms-2
the accelartion due to gravity =9.81ms-2)
F1= 225.63N
F2(force of friend)=21*9.81ms-2
F2 = 206.01N
so distance friend needs to be(d2)
d2= (F1*d1)/F2
d2=(225.63*1.5)/206.01 = 1.701m
got the explanation and answer!!!!!
2007-03-15 14:57:50 · answer #2 · answered by curious 2 · 0⤊ 0⤋
1.643 m
2007-03-15 14:56:11 · answer #3 · answered by p v 4 · 0⤊ 0⤋