Question

a uniform plank PQ of length 10m and mass (m)kg rests on 2 supports A and B, where /PA/=/BQ/=1m. A load of mass 8kg is placed on the plank at point C such that /AC/ = 3.5m. If the reaction is 100N, calculate the (i) value of m (ii)reaction at A(take g=10ms^-2)

Answer 1

Make a drawing:

Since the plank does not move, all forces and torques are balanced. Forces:

$$R_A+R_B-m_Cg-m_pg=0,\\$$

torques (consider A as the pivot):

$$\frac{m_p}{10}g\cdot0.5-\frac{9}{10}m_pg\cdot4.5-m_Cg\cdot3.5+100\cdot8=0,$$

the solution is

$$m_p=13\text{ kg},\\ R_A=110\text{ N}.$$