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Schach
3 months ago
14

A 4.5-m-long wooden board with a 24-kg mass is supported in two places. One support is directly under the center of the board, a

nd the other is at one end. What can be said about the forces exerted by the supports?

Physics
2 answers:
Ostrovityanka [3.2K]3 months ago
8 0

All the weight of the wooden board rests solely on the support situated at the center of the rod, while the other support positioned at one end experiences no reaction force, resulting in a 0 reaction force.

Thus, the reaction force at the central support corresponds to the weight of the board, whereas the end support has 0 reaction force.

inna [3.1K]3 months ago
4 0

Answer:

The force applied by the support at the board's center pushes upward and matches the board's weight, which is mg ( 24x9.81)N, totaling 235.44N.

At the opposite end, the support does not contribute any force since the board is already stabilized by the central support.

Explanation:

An elaborate explanation and calculation can be found in the image below

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(a) W=-19.25J

(b) W=-52.8J

Clarification:

Greetings.

(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:

W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J

This value is negative as it expands against the given pressure.

(b) Furthermore, if the process is conducted reversibly, the pressure might change, hence, we need to calculate the work using:

W=nRTln(\frac{V_1}{V_2} )

The moles are calculated based on the provided mass of argon:

n=6.56g*\frac{1mol}{39.95g}=0.164mol

Consequently, the work amounts to:

W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J

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