Question

the diagram shows a uniform metre rule. the rule is pivoted at its mid - point. a weight of 4.0n is suspended from the rule at the 5cm mark. the rule is held by a string at the 30cm mark. the rule is in equilibrium.
what is the upward force that the string exerts on the rule?
a 0.67n b 4.0n c 6.0n d 9.0n

Explanation:

Step1: Calculate the moment of the \(4.0N\) weight

The distance of the \(4.0N\) weight from the pivot is \(50 - 5=45cm\).
The moment \(M_1\) is \(M_1 = 4.0\times45\) (using the formula \(M = F\times d\), where \(F\) is the force and \(d\) is the perpendicular distance from the pivot).

Step2: Calculate the distance of the string from the pivot

The distance of the string from the pivot is \(50 - 30 = 20cm\). Let the force exerted by the string be \(F\). The moment \(M_2\) of the string's force about the pivot is \(M_2=F\times20\).

Step3: Use the principle of moments (\(M_1 = M_2\))

Since the rule is in equilibrium, \(4.0\times45=F\times20\).
Solve for \(F\): \(F=\frac{4.0\times45}{20}\).

$$F=\frac{180}{20}=9.0N$$

Answer:

D. \(9.0N\)