With the center of mass of the seesaw 0.160M to the left of the pivot (on the si

Question

With the center of mass of the seesaw 0.160M to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the problem solving strategy for static equilibrum.

The two children shown in Figure 9.9 are balanced on a seesaw of negligible mass. (This assumption is made to keep the example simple—more involved examples will follow.) The first child has a mass of 26.0 kg and sits 1.60 m from the pivot.(a) If the second child has a mass of 32.0 kg, how far is she from the pivot? (b) What is Fp , the supporting force exerted by the pivot?

Explanation / Answer

here,

mass of child 1 , m1 = 26 kg

mass of child 2 , m2 = 32 kg

x1 = 1.6 m

let the distance of seccond child from the Pivot be x2

(a)

taking moment of force about the Pivot

m1*g*x1 - m2 * g * x2 = 0

26 * 1.6 - 32 * x2 = 0

x2 = 1.3 m

she is 1.3 m from the Pivot

(b)

let the supporting force exerted by the pivot be Fp

Fp - m1*g - m2*g = 0

Fp - 26*9.8 - 32*9.8 = 0

Fp = 568.4 N

the supporting force exerted by the pivot is 568.4 N

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