MHB Muffin's GCSE Additional Maths Moments Question (from YAnswers)

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The discussion revolves around calculating the reaction forces at supports C and E for a uniform plank with two painters standing on it. The total load, including the weights of the painters and the plank, is 160g, equating to 1569.6 N. The user initially calculated the reactions as 1053.5 N and 514.5 N, but these values were questioned due to discrepancies with the answer book. A correct approach involves setting up simultaneous equations based on equilibrium conditions, leading to the conclusion that the reaction at support C should be 100g N. The calculations highlight the importance of using proper equilibrium equations and moment analysis to find accurate results.
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Two painters of mass 50kg and 70 kg stand on a uniform horizontal plank ABCDEFG of mass 40kg and length 8m.

B, C, D and E are respectively 1m, 2m, 5m and 6m from A. The 50kg painter stands at B, the 70kg painter at D and the plank's supported at C and E.

Find the reaction at each support I got the answers to be 1053.5 and 514.5 but my answer book tells me otherwise (although it has been wrong before) can someone please check this and explain where I've gone wrong please?

CB
 
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CaptainBlack said:
Two painters of mass 50kg and 70 kg stand on a uniform horizontal plank ABCDEFG of mass 40kg and length 8m.

B, C, D and E are respectively 1m, 2m, 5m and 6m from A. The 50kg painter stands at B, the 70kg painter at D and the plank's supported at C and E.

Find the reaction at each support I got the answers to be 1053.5 and 514.5 but my answer book tells me otherwise (although it has been wrong before) can someone please check this and explain where I've gone wrong please?

CB

Since the reaction forces must sum to the total load (the two painters and the weight of the plank acting at the centre of mass of the plank) which is \(160g = 1569.6 \, {\rm{N}}\) (taking \(g\) to be \(9.81\, {\rm{m/s^2}}\) ) your answer is not impossible so far (using \(g=9.8\, {\rm{m/s^2}}\) we have exact agreement).

You need to set up a pair of simultaneous equations under the assumption that the system is in equilibrium, the first is that the reaction forces sum to the load forces:

\(R_C+R_E = 160 g \)

The second is obtained by taking moments about some convenient point, as the system is in equilibrium these must sum to zero. A convenient point (as it eliminates one of the unknowns) is either C or E.

Taking moments about E:

\(4 R_C -2 (40 g) -5 (50 g) - 1 (70 g) =0\)

(the load forces give anti-clockwise moments and the reaction a clockwise moment, I am taking clockwise as positive here)

So:

\(R_C = 100 g\, {\rm{N}}\) ...

CB
 
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