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postnew [5]
3 years ago
12

A uniform beam of arbitrary, unsymmetrical cross-section and length 2???? is built-in at one end and simply supported in the ver

tical direction at a point halfway along its length. This support, however, allows the beam to deflect freely in the horizontal x direction. For a vertical load W applied at the free end of the beam, calculate bending moments for the sections at A and B.
Physics
1 answer:
leonid [27]3 years ago
5 0

Answer:

 the bending moment will be W from either sides

Explanation:

bending moment= force (load) * perpendicular distance, if I understand the question the distance will be 1/2 of the length

=> f x 1/2(l) =W*1/2(2) =W

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Answer: F_{2}=\frac{3}{4}F_{1}

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{m_{1}m_{2}}{r^2}

Where:

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G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

In this case we have two situations:

1) Two bags with masses 4M and 4M mutually exerting a gravitational attraction F_{1} on each other:

F_{1}=G\frac{(4M)(4M)}{r^2}   (1)

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2) Two bags with masses 2M and 6M mutually exerting a gravitational attraction F_{2} on each other (assuming the distance between both bags is the same as situation 1):

F_{2}=G\frac{(2M)(6M)}{r^2}   (4)

F_{2}=G\frac{12M^2}{r^2}   (5)

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Now, if we isolate \frac{GM^2}{r^2} from (3):

\frac{F_{1}}{16}=\frac{GM^2}{r^2}   (7)

Substituting \frac{GM^2}{r^2}  found in (7) in (6):

F_{2}=12(\frac{F_{1}}{16})   (8)

F_{2}=\frac{12}{16}F_{1}   (9)

Simplifying, we finally get the expression for F_{2}  in terms of F_{1} :

F_{2}=\frac{3}{4}F_{1}  

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