Question

find the component of a force of 100N in direction inclined to it at 30 degree and 40 degree on opposite side​

Answer 1

Given :

F = 100N

θ1 = 30°

θ2 = 40°

To find :

$F_x$ =?

$F_y$ =?

Step-by-step explanation:

Force 100N is inclined at an angle of 30° & 40° with the X-axis.

Components of force are Fx & Fy.

from fig.

1)

x component of force 100N of angle 30° is given by:

Fx = Fcosθ

    = 100 cos 30

Fx = 86.6 N

y component of force 100N of angle 30° is given by

Fy = Fsinθ

    = 100 sin 30

Fy  = 50 N    

2)

x component of force 100N of angle 40° is given by:

Fx = Fcosθ

    = 100 cos 40

Fx = 76.6 N

x component of force 100N of angle 40° is given by:

Fy = Fsinθ

    = 100 sin 40

Fy = 64.27 N

Refer following attachment:

Answer 2

Answer:

P=100N , a=40° , b = 30°

p¹ = P* sin b / sin (a+b)

= 100*sin30°/sin(40+30)

=100*0.5/sin70°

= 50/0.939

=53.2N

p²= P*sina/sin(a+b)

=100*sin40°/sin(40°+30°)

=100*0.642/sin70°

=64.2*0.939

=68.4

ANSWER P¹= 53.2. P²= 68.4