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It looks like the integral is
where <em>D</em> is the set
<em>D</em> = {(<em>x</em>, <em>y</em>) : 0 ≤ <em>x</em> ≤ 1 and <em>x</em> ² ≤ <em>y</em> ≤ √<em>x</em>}
So we have
Answer:
(i) The total anticlockwise moment about O, is 6000 + 50 × OD
(ii) The total clockwise moment about O, is 5000 + 50 × OC
(iii) The difference between the anticlockwise and the clockwise moment is 1000 N·cm
(iv) The distance from O where a 100 gf force should be placed to balance the meter rule is 10 cm to the right of the point O on the meter rule
Step-by-step explanation:
The given parameters are;
The mass of the uniform meter rule = 100 gf
The center of the meter rule = O
The mass of the weight at point A = 150 gf
The mass of the weight at point B = 250 gf
The distance of the point A from the point O = OA = 40 cm
The distance of the point B from the point O = OB = 20 cm
(i) Whereby the point A is on the left of the meter rule, and taking the mass of half the meter rule as acting on the center, of the midspan from the left edge to the center, O, which is the point, D we have;
The total anticlockwise moment, Mₐ = 150 × 40 + 50 × OD = 6000 + 50 × OD
(ii) The total clockwise moment about O, is given similarly as follows;
The total clockwise moment, = 250 × 20 + 50 × OC = 5000 + 50 × OC
(iii) The difference between the anticlockwise and the clockwise moment = Mₐ - = 6000 + 50 × OD - 5000 + 50 × OC
Where, OD = OC, we have;
Mₐ - = 6000 + 50 × OD - 5000 + 50 × OD = 1,000
Mₐ - = 1000 N·cm
Therefore, Mₐ >
(iv) The distance from O where a 100 g force should be placed to balance the meter rule should give a 1000 N·cm clockwise moment as follows;
Let the 100 gf load be placed at the point E, to the right of the point O, therefore;
100 gf × OE = 1000 N·cm
OE = 1000/(100) = 10
OE = 10 cm
Therefore, the 100 gf should be placed 10 cm to the right of the point O.