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miskamm [114]
3 years ago
12

Two pans of a balance are 46.3 cm apart. The fulcrum of the balance has been shifted 0.633 away from the center by a dishonest s

hopkeeper.
By what percentage is the true weight of the goods being marked up by the shopkeeper?
Answer in units of %
Physics
1 answer:
soldier1979 [14.2K]3 years ago
5 0

distance of each pan from the center or fulcrum is given as

r = 23.15 cm

now if dishonest shopkeeper shifted it by 0.633 cm from center

so distance on each side is given as

d_1 = 23.15 - 0.633 = 22.52 cm

d_2 = 23.15 + 0.633 = 23.78 cm

now the weight is balance as

W_1d_1 = W_2d_2

W(22.52) = W_2(23.78)

now we will have

W_2 = 0.95W

now we can find the percentage change as

percentage = \frac{W - W_2}{W} \times 100

percentage = 5%

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Answer:

μsmín = 0.1

Explanation:

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       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

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  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
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       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

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     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

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