Two forces 5N and 10N are acting at "O" and "P" respectively on a uniform rod of length 100 cm suspended at the position of cent
er of gravity 50cm mark as shown in figure .
1 answer:
Answer:
Option C. 75 cm
Explanation:
To obtain the position of P, let us calculate the value of y as shown in the attached photo.
The value of y can be obtained as follow:
Anticlockwise moment = clockwise moment
Anticlockwise moment = 5 × 50
Anticlockwise moment = 250
Clockwise moment = y × 10
Anticlockwise moment = clockwise moment
250 = y × 10
Divide both side by 10
y = 250/10
y = 25 cm
Finally, we shall obtain the value of P as follow:
Since P lies after the pivot (i.e 50 cm), therefore,
P = 50 + y
y = 25
P = 50 + 25
P = 75 cm
Therefore, the position of P on the metre rod is 75 cm
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Answer:
F = 36 kN
Explanation:
It is given that,
The pressure on the base of the fish tank is 4000N/m².
The base of the tank is a rectangle measuring 2.0m by 4.5m.
Area of the base of the tank is 9 m²
We need to find the force on the base caused by the base of the water. Pressure on the base of the tank is given by the force acting per unit area such that,
So, the force of 36 kN is acting on the base by the base of water.