Detrmine the resultant force?

Question

F1 is 600N, 70degrees..F2 is 400 N, 20 degrees...and F3 is 500N, 60 degrees.

Answer 1

You need to calculate the horizontal components of each force and the vertical components of each force, using trigonometry. Then add them together, separately, to get a new horizontal component and a new vertical components. Then, using trigonometry again, determine the resultant that corresponds to the summed components.

Looking at just the first force, 600N at 70 degrees (from the horizontal, I assume), the horizontal component is 600*cos(70) = 205.2 N and the vertical component is 600*sin(70) = 563.8 N.

Now, you're going to calculate three horizontal components and three vertical components, and then sum them to get a new horizontal component and a new vertical component. But to show you how it's done, I'll use the single components I just calculated. The magnitude of the resultant is the square root of the sum of the squares of the components; that is, it's the length of the hypotenuse of the right triangle formed by the components. That's sqrt(205.2^2 + 563.8^2) = 600.0. The angle is equal to the arctangent of the vertical component divide by the horizontal component, or atan(563.8/205.2) = 70.0. So you can see the method works, and you can use it for the larger problem.

Answer 2

I've replied to this question in the engineering section. There is some information missing if you want to get the answer that you quote there.

If however, we assume that all the angles are measured with respect to the positive x-axis then we can break the forces down as follows

F1_x = 600Cos70 = 205.2 N
F1_y = 600Sin70 = 563.8 N

F2_x = 400Cos20 = 375.9 N
F2_y = 400Sin20 = 136.8 N

F3_x = 500Cos60 = 250 N
F3_y = 500Sin60 = 433.0 N

The net force in the x-direction is 831.1 N and in the y-direction is 1133.6 N

The magnitude of the net force is SQRT(831.1^2 + 1133.6^2)1406 N

The angle with the x-axis is arctan(1133.6/831.1) = 54 DEG