Question

What are (a) the x component and (b) the y component of a vector in the xy plane if its direction is 259° counterclockwise from the positive direction of the x axis and its magnitude is 5.4 m?

Answer 1

Answer:

(a) -1.030m

(b) -5.301m

Explanation:

Given a vector F in the xy plane, of magnitude F and in a direction θ counterclockwise from the positive direction of the x-axis;

The x-component ($F_{X}$) of vector F is given by;

$F_{X}$ = F cos θ    ---------------------(i)

And;

The y-component ($F_{Y}$) of vector F is given by;

$F_{Y}$ = F sin θ            -----------------------(ii)

Now to the question;

Let the vector be A

Therefore;

The magnitude of vector A is A = 5.4m

The direction θ of A counterclockwise from the positive direction of the x-axis = 259°

(a) The x-component ($A_{X}$) of the vector A is therefore given by;

$A_{X}$ = A cos θ       ------------------------(iii)

Substitute the values of θ and A into equation (iii) as follows;

$A_{X}$ = 5.4 cos 259°

$A_{X}$ = 5.4 x (-0.1908)

$A_{X}$ = -1.030

Therefore, the x-component of the vector is -1.030m

(b) The y-component ($A_{Y}$) of the vector A is therefore given by;

$A_{Y}$ = A sin θ       ------------------------(iv)

Substitute the values of θ and A into equation (iv) as follows;

$A_{Y}$ = 5.4 sin 259°

$A_{Y}$ = 5.4 x (-0.9816)

$A_{Y}$ = -5.301m

Therefore, the y-component of the vector is -5.301m

Answer 2

Answer:

|Ax| =1.03 m  (directed towards negative x-axis)

|Ay|= 5.30 (directed towards negative y-axis)

Explanation:

Let A is a vector = 5.4 m

θ = 259°

to Find Ax, Ay

Sol:

according the condition it lies in 3rd quadrant

we know that Horizontal Component Ax = A cos θ

Ax = 5.4 Cos 259°

Ax = - 1.03 m

|Ax| =1.03 m  (directed towards negative x-axis)

Now Ay = A sin θ

Ay = 5.4 Sin 259°

Ay = -5.30

|Ay|= 5.30 (directed towards negative y-axis)