A crowbar 1.5 m long rests on a stone 0.25 m from the end under a 900 N load…

Step 1. (a) $900\ \text{N} \times 0.25\ \text{m} = E \times 1.25\ \text{m}$, so $E = 180\ \text{N}$.

Step 2. (b) $\dfrac{\text{load}}{\text{effort}} = \dfrac{900\ \text{N}}{180\ \text{N}} = 5$.