A uniform metre rule of mass 150 g is pivoted freely at the 0cm mark what force applied vertically upwards at the 60cm mark is needed to maintain the rule horizontally
.15 * 9.8 * 50 = 60 * F ... Newtons
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To maintain the uniform meter rule horizontally, the sum of the moments about the pivot point (0 cm mark) should be zero.
First, let's calculate the clockwise and counterclockwise moments. The moment of a force is given by the formula:
Moment = Force x Distance
Clockwise moments are considered negative, while counterclockwise moments are considered positive.
In this case, the force is applied vertically upwards at the 60 cm mark. Given that the rule is pivoted at the 0 cm mark, the distance between the force and the pivot point is 60 cm.
Let's assume the force required to maintain the rule horizontally is F (in Newtons).
The clockwise moment is (-F) x (60 cm) = -60F for the force at the 60 cm mark.
Since there are no other forces mentioned in the problem, the total clockwise moment is zero.
Therefore, the sum of all moments about the pivot point is:
Counterclockwise moment + Clockwise moment = 0
(0 cm)(W) + (-60 cm)(F) = 0
Since the weight (W) of the meter rule is not given, we cannot solve for F specifically.
However, we know that the weight acts vertically downward, which means its moment is counterclockwise. The weight can be calculated using:
Weight = mass x acceleration due to gravity
Given that the mass is 150 g = 0.15 kg, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight:
Weight = (0.15 kg) x (9.8 m/s^2) = 1.47 N
Using this value for the weight (W), we can solve for F:
(0 cm)(1.47 N) + (-60 cm)(F) = 0
0 - 60F = 0
-60F = 0
Dividing both sides by -60:
F = 0 N
Hence, to maintain the meter rule horizontally, no force (0 N) is needed at the 60 cm mark when the weight of the meter rule is 1.47 N.