The two forces of 9 Newtons and 12 Newtons which are acting at ri

SOLUTION

Concept:

Methods of finding resultant of forces

A) Law of a parallelogram of forces

Where P and Q are two forces acting on a body with angle θ between them

α = Angle made by resultant with respect to force P

α is the direction of resultant R

\({\rm{R}} = \sqrt {{{\rm{P}}^2} + {{\rm{Q}}^2} + 2{\rm{PQ}}\cos {\rm{\theta }}} \)

\({\rm{\alpha }} = {\tan ^{ - 1}}\left[ {\frac{{{\rm{Q}}\sin {\rm{\theta }}}}{{{\rm{P}} + {\rm{Q}}\cos {\rm{\theta }}}}} \right]{\rm{\;\;}}\)

B) Method of resolution of forces

ΣF_X = summation of forces acting in X-direction

ΣF_Y = Summation of forces acting in Y-direction

\({\rm{R}} = \sqrt {{{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{X}}}} \right)}^2} + {{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{Y}}}} \right)}^2}} \)

\({\rm{\theta }} = {\tan ^{ - 1}}\left| {\frac{{{\rm{\Sigma }}{{\rm{F}}_{\rm{Y}}}}}{{{\rm{\Sigma }}{{\rm{F}}_{\rm{X}}}}}} \right|\)

Calculation:

Let the force P = 9 N and Q = 12 N and θ = 90°

By law of parallelogram of forces

\({\rm{R}} = \sqrt {{{\rm{P}}^2} + {{\rm{Q}}^2} + 2{\rm{PQ}}\cos {\rm{\theta }}} \)

\({\rm{R}} = \sqrt {{9^2} + {{12}^2} + 2 \times 9 \times 12 \times \cos 90^\circ } \)

\({\rm{R}} = \sqrt {{9^2} + {{12}^2}} \)

R = 15 N

Alternate method:

By the method of resolution

ΣF_X = 9 N, ΣF_Y = 12 N

\({\rm{R}} = \sqrt {{{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{X}}}} \right)}^2} + {{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{Y}}}} \right)}^2}} \)

\({\rm{R}} = \sqrt {{9^2} + {{12}^2}} \)

R = 15 N