Question

The resultant of two concurrent forces is
perpendicular to the smaller force and
angle between the forces is 120° if the
bigger force is 60 N, find the smaller
force is
40 N
30 N
20 N
10N​

Answer 1

Answer:

I think twenty newtons wills come see

Answer 2

$\Large\bf{\color{indigo}GiVeN,}$

✅ We solve this problem in vector format.

$\bf\pink{Let,}$

• $\bf{\vec{F}_1}$ is the smaller force vector.

• $\bf{\vec{F}_2}$ is the larger force vector.

• The resultant force vector $\bf{\vec{F}_R}$ is making an angle 90° with the smaller force $\bf{\vec{F}_1}$.

☆ The resultant force vector $\bf{\vec{F}_R}$ makes an angle,

⇒ α = 120° - 90° = 30°, with the larger force vector $\bf{\vec{F}_2}$, as shown in the diagram.

$\Large\bf{\color{coral}To\:FiNd,}$

• The smaller force.

$\Large\bf{\color{lime}CaLcUlAtIoN,}$

$\bf\blue{From\:right\:angle\:\triangle{ADC},}$

$:\implies\:\:\bf{\sin{\alpha}\:=\:\dfrac{Opposite\:side}{Adjacent\:side}\:}$

$:\implies\:\:\bf{\sin{\alpha}\:=\:\dfrac{CD}{AC}\:}$

$\bf\red{━─━─━─━─━─━─━─━─━─━─━─━─━─━─━─}$

$\longrightarrow\:\:\bf{AC\:=\:magnitude\:of\:\vec{F}_2\:=\:60N}$

$\longrightarrow\:\:\bf{CD\:=\:AB\:=\:magnitude\:of\:\vec{F}_1}$

$\bf\red{━─━─━─━─━─━─━─━─━─━─━─━─━─━─━─}$

$:\implies\:\:\bf{\sin{\alpha}\:=\:\dfrac{\mid\vec{F}_1\mid}{\mid\vec{F}_2\mid}\:}$

$:\implies\:\:\bf{\sin{30°}\:=\:\dfrac{\mid\vec{F}_1\mid}{\mid\vec{F}_2\mid}\:}$

$:\implies\:\:\bf{\mid\vec{F}_1\mid\:=\:\mid\vec{F}_2\mid\times{\sin{30°}}}$

$:\implies\:\:\bf{\mid\vec{F}_1\mid\:=\:60\times{\dfrac{1}{2}}}$

$:\implies\:\:\bf\purple{\mid\vec{F}_1\mid\:=\:30\:N}$

∴ (b) The smaller force is 30 N.