Component method vector problem

[ur5pointos2sl]

Use the component method to find the resultant(direction and magnitude) of the three forces shown.

Could someone please verify my answers or tell me where I went wrong.

Ax = 500 Cos 30 = -433N
Ay = 500 Sin 30 = -250N
Bx=-250 N
By= 0
Cx=200 Cos 20 = 187.9N
Cy= 200 Sin 20 = 68.4N

R = (-433-250+187.9)I + (-250 + 0 + 68.4)j
= (-495.1, -181.6)
R = sqrt(-495.1^2 + -181.6^2)
= 527N

Tan Theta = 318.4 / 370.9
Theta = 20.1 deg.

 

[Redbelly98]

Use the component method to find the resultant(direction and magnitude) of the three forces shown.

Could someone please verify my answers or tell me where I went wrong.

Ax = 500 Cos 30 = -433N
Ay = 500 Sin 30 = -250N
Bx=-250 N
By= 0
Cx=200 Cos 20 = 187.9N
Cy= 200 Sin 20 = 68.4N

R = (-433-250+187.9)I + (-250 + 0 + 68.4)j
= (-495.1, -181.6)
R = sqrt(-495.1^2 + -181.6^2)
= 527N

That seems right.

Tan Theta = 318.4 / 370.9
Theta = 20.1 deg.

Looks like the "318" and "370" are typos, and that you actually (and correctly) used 181.6/495.1 instead.

You'll also need to specify which half-axis that angle is measured from (+x, +y, -x, or -y), and whether the angle is above/below/right/left of that half-axis.

If you don't specify that, people normally assume it's from the +x axis, in the counterclockwise direction.

 

[tomcue]

Your calculations and method appear to be correct. To verify, you can use the Pythagorean theorem to find the magnitude of the resultant force:

R = sqrt((-495.1)^2 + (-181.6)^2) = 527N

To find the direction of the resultant force, you can use the inverse tangent function:

Theta = arctan(318.4/370.9) = 20.1 degrees

Therefore, the resultant force has a magnitude of 527N and a direction of 20.1 degrees, as you have calculated. Great job!