Answer:
(a) 0.42 m/s²
(b) 83 N
Explanation:
Assuming g = 10 m/s².
Draw free body diagrams for the child and the chair.
There are three forces on the child:
Tension force T pulling up,
Weight force Mg pulling down,
and normal force N pushing up.
There are three forces on the chair:
Tension force T pulling up,
Weight force mg pulling down,
and normal force N pushing down.
Sum of forces on the child in the y direction:
∑F = ma
T + N − Mg = Ma
Sum of forces on the chair in the y direction:
∑F = ma
T − N − mg = ma
Add the equations together to find the acceleration.
2T − Mg − mg = Ma + ma
a = (2T − Mg − mg) / (M + m)
a = g (2T − Mg − mg) / (Mg + mg)
a = (10 m/s²) (2 × 250 N − 320 N − 160 N) / (320 N + 160 N)
a = 0.42 m/s²
Plug into either equation to find the normal force.
T + N − Mg = Ma
250 N + N − 320 N = (320 N / 10 m/s²) (0.42 m/s²)
N = 83 N
