Question

a uniform beam xy is 100 cm long and weighs 4.0 n.
the beam rests on a pivot 60 cm from end x.
a load of 8.0 n hangs from the beam 10 cm from end x.
the beam is kept balanced by a force f acting on the beam 80 cm from end x.
what is the magnitude of force f?
a 8.0 n b 18 n c 22 n d 44 n

Explanation:

Step1: Calculate the moment of the 8.0N force

The distance of the 8.0N force from the pivot is \(60 - 10=50\space cm = 0.5\space m\).
The moment \(M_1\) is \(M_1 = 8.0\times0.5\) (using \(M = F\times d\), where \(F\) is force and \(d\) is perpendicular distance from pivot).
\(M_1=4.0\space N\cdot m\) (clock - wise moment).

Step2: Calculate the moment of the beam's weight

The beam's weight \(W = 4.0\space N\). The distance of the center of the beam from the pivot is \(60-(100\div2)=10\space cm=0.1\space m\).
The moment \(M_2\) of the beam's weight is \(M_2=4.0\times0.1\) (clock - wise moment).
\(M_2 = 0.4\space N\cdot m\).

Step3: Calculate the distance of force \(F\) from the pivot

The distance of force \(F\) from the pivot is \(80 - 60=20\space cm = 0.2\space m\). Let the moment of force \(F\) be \(M_3\). Using the principle of moments (\(\sum M_{clock - wise}=\sum M_{anti - clock - wise}\)), and \(M_3=F\times0.2\) (anti - clock - wise moment).
\(M_1 + M_2=M_3\).
Substitute \(M_1 = 4.0\space N\cdot m\) and \(M_2 = 0.4\space N\cdot m\) into the equation: \(4.0+0.4=F\times0.2\).

Step4: Solve for \(F\)

First, simplify the left - hand side: \(4.4=F\times0.2\). Then, solve for \(F\) using \(F=\frac{4.4}{0.2}\).

Answer:

\(F = 22\space N\), so the answer is C.