At the city playground, you observe a 35 - pound child sitting on the right end

Question

At the city playground, you observe a 35 - pound child sitting on the right end of a horizontal see - saw (teeter - totter), 1.5 m from the pivot point. On the other (left) side of the pivot the child's mother pushes straight down on the see - saw with a force of 95 N. In which direction (CW or CCW, from your vantage point) and with what angular acceleration, a, does the see - saw rotate if the mother applies the force at a distance of (a) 3.0 m, (b) 2.5 m, or (c) 2.0 m from the pivot? The see - saw board has a mass of 15 kg and a length, L = 6.0 m. Assume the child is a point mass (I = mr^2) and the see -saw board can be treated as along thin uniform rod (1 = 1/12 ML^2).

Explanation / Answer

Given data.

Mass of child = 35 pounds= 15.875 kg

Force applied by mother = 95 N

MOI of see saw board = (ml^2)/12 = 15 × 6×6/12 = 45 kg-m^2

MOI of child = m r^2 = 15.875 ×3 × 3 = 142.875 kg-m^2

Part a) when force is applied at 3 m

Write torque equation about pivot of seesaw

15.875x9.81 N x 3m - 95Nx 3m =(45+142.875)x a

a= .97 rad/s2 cw

b)15.875x9.81 N x 3m - 95Nx 2.5m =(45+142.875)x a

a= 1.22 rad/s2 cw

c)15.875x9.81 N x 3m - 95Nx 2m =(45+142.875)x a

a=1.475 rad/s2 cw

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