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babymother [125]

2 years ago

11

A non-uniform rod 60cm long and weighs 32N is balanced at the 45cm mark. A load of 2N is hung on the zinc rod at the 25cm mark.

Where must a second knife-edge be placed to balance the zinc rod horizontally

1 answer:

Pavlova-9 [17]2 years ago

7 0

Answer:

The second knife-edge must be placed 46.2 cm from the zero mark of the rod.

Explanation:

From the law of equilibrium, ΣF = 0 and ΣM = 0.

Let R be the reaction at the knife edge. Since the weight of the rod and zinc load act downward, and we take downward position as negative

-32 N - 2 N + R = 0

-34 N = -R

R = 34 N

Also, let us assume the knife-edge is x cm from the zero mark. Taking moments about the weight and assuming the knife-edge is right of the weight of the rod. Taking clockwise moments as positive and anti-clockwise moments as negative,

-(45 - 25)2 + (x - 45)R = 0

-(20)2 + (x - 45)34 = 0

-40 = -(x - 45)34

x - 45 = 40/34

x - 45 = 1.18

x = 45 + 1.18

x = 46.18 cm

x ≅ 46.2 cm

The second knife-edge must be placed 46.2 cm from the zero mark of the rod.

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We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

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where again, we must be careful to the signs of the various quantities:

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Substituting t = 2.64 s, we find the final velocity of the stone:

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