When the 0.08-kg body is in the position shown, the linear spring is stretched 8 mm. Determine the force P required to brea at C. Complete solutions for (a) including the effects of the weight and (b) neglecting the weight. P 64 mm 64 mm k = 1550 N/m B 36 mm 36 mm

Given: The mass of the body is 0.08 kg. 

           The spring is stretched to 8 mm. 

           The spring constant is 1550 N/m 

To determine: (a) The force P when weight is included 

Note: 1 mm is equal to 0.001 m, so 8 mm is equal to 0.008 m. 

The formula for spring force is 

$\mathrm{F}=\mathrm{kx}$

where k is the spring constant and x is the stretch on spring 

Substitute the values 

$$\begin{gathered} \mathrm{F}=1550 \mathrm{N}/\mathrm{m}\times 0.008 \mathrm{m} \\ \mathrm{F}=12.4 \mathrm{N} \end{gathered}$$

The moment is equal to the product of force and perpendicular distance to the pivot point. 

The anticlockwise direction is taken as positive and clockwise direction is taken as negative.