Question 3.21: A roller shown in Fig. 3.32 is of mass 150 kg. What force P ...

Consider the free body diagram of roller, shown in Fig. 3.32 (a). The forces acting on roller will be P, N, weight of roller and reaction of block (R_{A}) when roller tends to move over the block, its contact from the floor terminates and thus reaction of the floor also disappears

i.e., \qquad \qquad N = 0

Thus only three forces kept the roller in equilibrium thus all forces have to be concurrent, that’s why reaction of block R_{A} passes through the center of the roller.

Let                \angle A O B=\theta, \text { In } \triangle A O B,

\cos \theta=\frac{O B}{O A}=\frac{175-100}{175}

θ = 64.62°

Considering forces about point O, further we can use Lami’s theorem or Resolution method.
Using Lami’s theorem,

\frac{R_{A}}{\sin \left(90^{\circ}+25^{\circ}\right)}=\frac{P}{\sin \left(180^{\circ}-\theta\right)}=\frac{150 \times g}{\sin \left(\theta+90^{\circ}-25^{\circ}\right)}

P=\frac{150 g \sin \theta}{\sin \left(\theta+65^{\circ}\right)}

Substituting value of θ and g = 9.81 m/sec²

P=\frac{150 g \sin 64.62^{\circ}}{\sin \left(64.62^{\circ}+65^{\circ}\right)}

P = 1725.94 N

Using Moment method,
Taking moment about A, as we need only P,

P \cdot \cos 25^{\circ} \times O B+P \cdot \sin 25^{\circ} \times A B=150 \times 9.81 \times A B

P\left[75 . \cos 25^{\circ}+158.11 \sin 25^{\circ}\right]=150 \times 9.81 \times 158.11

OB = 175 – 100

=75 mm

A B=\sqrt{175^{2}-75^{2}}

= 158.11 mm

P=\frac{150 \times 9.81 \times 158.11}{\left(75 . \cos 25^{\circ}+158.11 \sin 25^{\circ}\right)}

P = 1726.04 N