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Misha Larkins [42]
3 years ago
11

Five identical keys are suspended from a balance, which is held horizontally as shown. The two keys on the left are attached to

the balance 6 cm from the pivot and the three keys on the right are attached 5 cm from the pivot. What will happen when the person lets go of the balance beam?
Engineering
1 answer:
Ann [662]3 years ago
6 0

Answer:

movement in clockwise direction.

Explanation:

The following parameters or information are given from the question above, they are:

[1]. There are two identical keys, [2]. two out of the five keys are attached to 6cm from the pivot, [3]. three keys out of the five keys on the right are attached 5 cm.

Therefore, considering the moment of force, the two keys on the left = 2 × 6 = 12.

Also, considering the moment of force, the 3 keys on the right = 3 × 5 = 15.

Therefore, we have more weight on the right keys. So, in order to balance the force there must be movement in clockwise direction.

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Given data:

width= 25 mm = 25\times 10^{-3} m

thickness = 6.5 mm = 6.5\times 10^{-3} m

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\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}

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we know that

k =\sigma_{applied} (\sqrt{\pi C})

  =6.154\sqrt{\pi (2.5\times 10^{-04})}          [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

ifK_C = 1.15 Mpa m^{1/2} then load will be

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\sigma _{frac} = 41.04 MPa

load = \sigma _{frac}\times Area

load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N

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