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5 votes
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?


B) If the actual counterweight fitted to this boom was 2350N: i) what additional downward-example-1
User Parthiban
by
5.8k points

2 Answers

1 vote

Final answer:

The question pertains to calculating torque and forces acting on a boom with a counterweight. To find the additional force applied by an attendant and the normal reaction on the boom, the principles of static equilibrium are used. However, without the system's geometry or the position of forces, numerical answers cannot be provided, only the approach.

Step-by-step explanation:

The question involves the application of static equilibrium principles to a physical situation involving a boom and forces exerted on it. To solve this problem, one must use concepts of torque and force balance. When a counterweight of 2350 N is fitted to the boom, the additional downward force that must be applied by the car park attendant can be found by equating torques around the pivot or by summing forces in the vertical direction. Likewise, the normal reaction at the end support when the boom is resting on it requires summing all the vertical forces including the weight of the boom and the counterweight, and accounting for the force applied by the attendant.

Suppose x is the additional downward force that the attendant must apply. Then the total downward force, which includes the boom's weight (3000 N) and the counterweight (2350 N), should be balanced by the upward normal force at the support and the force applied by the attendant.

Therefore, the normal reaction, N, can be calculated as:

N = 3000 N (weight of the boom) + 2350 N (counterweight) - x (additional force applied by the attendant).

Since details on the system's geometry or the position of the attendant applying the force are not provided, we cannot give numerical answers to i) and ii), but the approach to solving such a problem has been outlined.

User Kev
by
6.5k points
3 votes

Answer:

50 N

4.2 N

Step-by-step 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

User JYL
by
6.0k points