Answer:
2.4 meters
Explanation:
Let
m1= 10 kg
m2= 6 kg
Distance of m2 from pivot = 4 m
Distance of m1 from pivot = x (suppose) =?
If system is not rotating then it should be in equilibrrium position. And in this case clock-wise and anti -clock wise torques should balance each other i-e
weight of m1 mass bolck × x = weight of m2 mass block × 4
m1×g× x = m2 × g × 4
==> 10×9.8×x= 6×9.8×4
==> x = 6×9.8×4 / 10×9.8
==> x = 24/10=2.4 m
If the 6.0 kg mass is 4.0 m away from the pivot, the 10kg mass is far away from the pivot by 2.4 m
Moment: Moment of a force is the product of force and its perpendicular distance. The unit of moment is Nm
In other to solve this question, we need to make use of the principle of moment.
Principle of moment: For a body to be at equilibrium, the sum of clockwise moment = the sum of anticlock wise moment.
For the system to be at equilibrium (not rotating),
The sum of clockwise moment must be equal to the sum of anticlock wise moment. (Principle of moment)
From the question,
moment = mass×garvity×distance
M = mgd............ Equation 1
where m = mass, d = distance g = gravity = 10 m/s²
Clockwise moment = 6×10×4 = 240 Nm.
Anticlock wise moment = 10×10×y = 100y
Where y = distance of the 10 kg mass from the point of pivot
therefore,
240 = 100y
y = 240/100
y = 2.4 m.
Hence, the 10 kg mass is 2.4 m away from the pivot
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