A large fish hangs from a spring balance supported from the roof of an elevator.

Question

If the elevator has an upward acceleration of 2.85m/s^2 and the balance reads 60.0N , what is the true weight of the fish?
Under what circumstances will the balance read 39.0N ?

Answer 1

True Weight of fish:

F= ma + mg

Where:
F=60.0N
a=2.85 m/s^s
g=9.80 m/s^2

Plug into formula where you are solving for m:

F = ma + mg

60.0 N = (2.85)m + (9.80)m
60.0 N = (12.65)m
m= 60.0N/12.65
m= 4.74 kg => 4740 g => 46.5 N

Under what circumstances will the balance read 39.0 N?

Formula: ∑F = F(normal) + F(weight)

Where:
∑F= a(4.74kg)
F(normal)= 39.0 N
F(weight)=(-9.80m/s^2)(4.74kg)

Plug into formula:

∑F = F(normal) + F(weight)
a(4.74) = 39.0 N + (-9.80)(4.74)
a(4.74) = -7.452
a = -1.57 m/s^2

Answer 2

thanks