Question

Find the component of a 50 N force which makes an angle of 45° with z-axis and whose projection in XY-plane makes 45° with the X-axis.
a) Fx = 25 N
b) Fy = 25√2 N
c) Fy= 25 N
d) Fz =25√2 N​

Answer 1

Answer:

We know that

F= acos theta

Fz= 50 cos45

=50*1 by root 2

rationalise the value of. cos 45

then we get ,Fz=25 root 2N

Answer 2

Answer:

The correct answer is option a) Fₓ = 25 N.

Concept:

Resolution of a vector along the axes:

Suppose a vector F has the coordinates as (x, y, z).

Its projection in XY plane is given by,

$F_{xy}$ = F sin θ

Its projection in the z direction is given by,

$F_z$ = F cos θ

Where θ is the angle between force and x-axis or z-axis.

Given:

Force, F = 50 N

Angle between force and z-axis, θ = 45°

Angle between the projection of force in XY plane and x-axis, Ф = 45°

Find:

Component of force.

Solution:

Given that, F = 50 N

Its component along z-axis is given by,

$F_z$ = F cos θ

$F_z$ = 50 cos 45°

$F_z$ =  50 × (1/√2)

$F_z$ = 50/√2

$F_z$ = 25√2 N

Its component along XY plane is given by,

$F_{xy}$ = F sin θ

$F_{xy}$ = 50 sin 45°

$F_{xy}$ = 50 × (1/√2)

$F_{xy}$ = 50/√2

$F_{xy}$ = 25√2 N

Its component along x-axis is given by,

$F_x$ = $F_{xy}$ cos Ф

$F_x$ = 25√2 cos 45°

$F_x$ = 25√2 × 1/√2

$F_x$ = 25 N

Its component along y-axis is given by,

$F_y$ = $F_{xy}$ sin Ф

$F_y$ = 25√2 sin 45°

$F_y$ = 25√2 × 1/√2

$F_y$ = 25 N

Hence, the component of a 50 N force which makes an angle of 45° with z-axis and whose projection in XY-plane makes 45° with the X-axis is Fₓ = 25 N.

So, option a) is the correct answer.

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