Question

A force is inclined at 50° to the horizontal. If its rectangular component in the horizontal direction be 50 N, find the magnitude of the force and its vertical component.

Answer 1

Answer:

Let the force be F

it is inclined at 50 degree with horizontal.

So F cos50 = 50N

⇒ F = 50/cos50 N

⇒ F = 77.78 N

vertical component, F sin50 = 77.78×sin50 = 59.58 N

Answer 2

$\huge {\rm {\underline {Question}}}$

A force inclined at 50 ° to horizontal , if it's component in this direction is 50 N , find magnitude of force and its vertical components.

$\huge {\rm {\underline {Answer}}}$

Here $\tt {F_x = 50N}$ , $\theta = 50°$

But $\tt {F_x = F cos \theta}$

Therefore ,

$\bf {F = \frac {F_x}{cos \theta} = \frac {50}{cos 50°} = \frac {50}{0.6428} = 77.78N}$

Also ,

$F_y \\ \implies F sin \theta \\ \implies 77.78 \times sin50° \\ \implies 77.78 \times 0.7660 \\ \implies 59.58N$

Hope it helps dear....