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)
(a) -1.030m
(b) -5.301m
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 ([tex]F_{X}[/tex]) of vector F is given by;
[tex]F_{X}[/tex] = F cos θ ---------------------(i)
And;
The y-component ([tex]F_{Y}[/tex]) of vector F is given by;
[tex]F_{Y}[/tex] = 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 ([tex]A_{X}[/tex]) of the vector A is therefore given by;
[tex]A_{X}[/tex] = A cos θ ------------------------(iii)
Substitute the values of θ and A into equation (iii) as follows;
[tex]A_{X}[/tex] = 5.4 cos 259°
[tex]A_{X}[/tex] = 5.4 x (-0.1908)
[tex]A_{X}[/tex] = -1.030
Therefore, the x-component of the vector is -1.030m
(b) The y-component ([tex]A_{Y}[/tex]) of the vector A is therefore given by;
[tex]A_{Y}[/tex] = A sin θ ------------------------(iv)
Substitute the values of θ and A into equation (iv) as follows;
[tex]A_{Y}[/tex] = 5.4 sin 259°
[tex]A_{Y}[/tex] = 5.4 x (-0.9816)
[tex]A_{Y}[/tex] = -5.301m
Therefore, the y-component of the vector is -5.301m
