A uniform beam, WX, of length 90 cm and weight 50N is suspended on a pivot, 35 cm from W. It is kept in equilibrum by a means of forces T and 20N applied at...
Question 1 Report
A uniform beam, WX, of length 90 cm and weight 50N is suspended on a pivot, 35 cm from W. It is kept in equilibrum by a means of forces T and 20N applied at Y and Z respectively. |WY| = 10cm and |XZ| = 10cm. Find the value of T
Answer Details
Set distances from end \(W\). The beam is \(90\,\text{cm}\) long.
- \(Y\): \(|WY| = 10\,\text{cm}\), so \(Y\) is at \(10\,\text{cm}\).
- Pivot: \(35\,\text{cm}\) from \(W\).
- Weight \(50\,\text{N}\) acts at the centre (uniform beam): \(45\,\text{cm}\).
- \(Z\): \(|XZ| = 10\,\text{cm}\), so \(Z\) is at \(90 - 10 = 80\,\text{cm}\).
Taking moments about the pivot (which removes the pivot reaction). Distances from the pivot:
- \(T\) at \(Y\): \(35 - 10 = 25\,\text{cm}\) on the \(W\) side.
- Weight \(50\,\text{N}\): \(45 - 35 = 10\,\text{cm}\) on the \(X\) side.
- \(20\,\text{N}\) at \(Z\): \(80 - 35 = 45\,\text{cm}\) on the \(X\) side.
For equilibrium, the anticlockwise moment of \(T\) balances the clockwise moments of the weight and the \(20\,\text{N}\) force:
\[T \times 25 = 50 \times 10 + 20 \times 45\] \[25\,T = 500 + 900 = 1400\] \[T = \frac{1400}{25} = 56\,\text{N}\]\(T = 56\,\text{N}\).
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