The seesaw in the figure below is 4.5 m long. Its mass of 20 kg is uniformly dis

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

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