B) if the actual counterweight fitted to this boom was
2350n:
i) what additional downward force must be applied
by the car park attendant to lift the boom
ii) what will be the normal reaction, n, at the end
support when the boom is resting on it?

50 N

4.2 N

Explanation:

i) The force needed to balance the boom is 2400 N.  If the weight of the counterbalance is 2350 N, then the downward force the park attendant must apply is 50 N.

ii) When the boom is resting on the end support, the normal force is:

∑τ = Iα

-W (0.50) + F (3.0) − N (6.0) = 0

-0.50 W + 3.0 F = 6.0 N

N = (-0.50 W + 3.0 F) / 6.0

N = (-0.50 × 2350 + 3.0 × 400) / 6.0

N ≈ 4.2

i think if it were me i whould divide 300 dide 15

step-by-step explanation: