The seesaw in the figure below is 4.5 m long. Its mass of 20 kg is uniformly dis
ID: 2240561 • Letter: T
Question
The seesaw in the figure below is 4.5 m long. Its mass of 20 kg is uniformly distributed. The child on the left end has a mass of 14 kg and is a distance of 1.4 m from the pivot point while a second child of mass 49 kg stands a distance L from the pivot point, keeping the seesaw at rest.
?F = = 0 ?? == 0 The seesaw in the figure below is 4.5 m long. Its mass of 20 kg is uniformly distributed. The child on the left end has a mass of 14 kg and is a distance of 1.4 m from the pivot point while a second child of mass 49 kg stands a distance L from the pivot point, keeping the seesaw at rest.
Explanation / Answer
a) the see saw to be in rest means the rotation of the system ( plank, two childs) is zero about the pivot point.
YES, the see saw will in equilibrium.
c) condition for transational equilibrium:
loads and reactions are along vertical, for the see saw to be in transatinal equilibrium, net forces in vertical direction should be zero.
condition is:
Fs- Fl-Fr-Fp=0 (convention; upward reactions are positive);
Fs= Fl+Fr+Fp.
condition for rotational equilibrium:
sum of moments about the pivot point should be zero.
CONVENTION ( clock wise moments are positive);
-l+p+r=0;
p+r=l.
note: (moment caused by the support is zero since the line of action of the support reaction is passing through the pivot point)
d) from the above reacion;
Fp * 0.85 ( (4.5/2)-1.4) + Fr * L = Fl * 1.4;
20 * 0.85 + 49 * L = 14 * 1.4;
=> 49 * L = 2.6;
=> L= 0.053 meters.
= 5.3 CM