Respuesta :
Answer:
20 kg
Explanation:
Assuming that the board remains horizontal with the unknown mass on it, then F = kx
If we add the vertical forces to zero, we have something like this
2F - (M + 10)g = 0
2F = (M + 10)g, next, divide both sides by 2
F = (M + 10)g/2
Since we were able to sum the moments at the right end of the board to zero, we then proceed to find the unknown mass M
To start, we say, Let the clockwise moment is positive, and so
F * 4 + F * 2 - Mg * 3.5 - 10 * g * 2 = 0
4F + 2F - 3.5Mg - 10 * 2 * g = 0
6F - 3.5Mg - 10 * 2 * g = 0,
Remember from above, we say that
F = M + 10)g/2, now, all we do is substitute it inside this equation
6 * (M + 10)g/2 - 3.5Mg - 10 * 2 * g = 0
3 * (M + 10)g - 3.5Mg - 10 * g * 2 = 0, divide all sides by g(so as to eliminate it)
3 (M + 10) - 3.5M - 10 * 2= 0
3M + 30 - 3.5M - 20 = 0
-0.5M + 10 = 0
0.5M = 10
M = 10/0.5
M = 20 kg
The mass of the block that was placed on the board is; M = 20 kg
Since the two supports are the same, then it means their forces will be the same.
Thus;
Force at left support = F
Force at right support = F
Now, we are told that a block of unknown mass is placed on the stiff board. This means that the sum of the weight of the board and and the block on it will be; (M + 10)g
where M is the mass of the block.
Now, from equilibrium we know that sum of upward forces is equal to sum of downward forces. Thus;
F + F = (M + 10)g
Thus;
2F = (M + 10)g
F = ¹/₂(M + 10)g
Now, taking moments about the right end gives;
(F × 4) + (F × 2) - (M × 3.5)g - (10 × 2)g = 0
6F - 3.5Mg - 20g = 0
Put ¹/₂(M + 10)g for F to get;
6(¹/₂(M + 10)g) - 3.5Mg - 20g = 0
Divide through by g to get;
3(M + 10) - 3.5M = 20
3M + 30 - 3.5M = 20
3.5M - 3M = 30 - 20
0.5M = 10
M = 10/0.5
M = 20 kg
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