39. (Taken from problem 5 at the end of Chapter 9) Look at the children balanced

Question

39. (Taken from problem 5 at the end of Chapter 9) Look at the children balanced on the see-saw in the figure below, ti= 1.60 m. The center mass of the seesaw is located 0.16 m to the heft of the pivot point. The seesaw board has a mass of 12.0 kg which produces a weight of (12.0 kg) (9.80 m/s^2) = W3. The lighter child has a mass of 26.0 kg (mi) and the heavier child has a mass of 32.0 kg (m^2). a) Find r2 the distance the larger child must be silting from the function point in order for the sew-saw to balance. (10 Points) Answer: 1.36 m b) Compute Fp. the load on the pivot. Answer 686 N

Explanation / Answer

Given m 1 = 26 kg

m 2 = 32 kg

r 1 = 1.6 m

Distance of center of mass from the center r = 0.16 m

Mass of the board m = 12 kg

In ceter of mass , Sum of moments of mass = 0

m1 g x (r 1 )+ m g x r - m2 g x ( r2 ) = 0

Here negative sign indicates that the two distances are opposite in direction .

or m1 x (r 1 ) + m r = m2 x ( r2 )

m1 x (r 1 )+ m r = m2 x ( r2 )

26 x (1.6) +12 x 0.16= 32 x ( r2)

43.52 = 32 x ( r2)

( r2) = 1.36 m

(B) Sum of lower forces = Sum of Upper forces

W 1+ W 2+ W3 = F p

   (m 1x g) +(m 2xg) + (m x g ) = F p

(26 x9.8) + (32 x 9.8) + (12 x 9.8) = F p

F p = 686 N

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