Two known forces F1 = 3.9 N at 76 o below the +x axis and F2 = 3.9 N at 35 o above the –x axis are acting on an object. (a) What is the magnitude of the third force F3 so that the acceleartion of the object is zero?
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Note: I initially misread the problem (I misplaced the hyphen on the second force and solved it as though it were positioned at 35 degrees above the "x-axis" rather than the "minus x" axis. Anyway, the strategy is the same but the actual solution will be different. Here is a revised version:
You can see* that the first force pulls down and to the right. The second force pulls up and to the left. We just need to split F1 and F2 into their horizontal "i" and vertical "j" components to determine the necessary components of F3.
F1 + F2 = 3.9 N * [ cos 76 I - sin 76 j ] + 3.9 N * [ - cos 35 i + sin 35 j ]
= 3.9 N [ (cos 76 - cos 35) i + (sin 35 - sin 76) j ]
Now, we need a third force such that F1 + F2 + F3 = 0 (i.e. F3 = - [ F1 + F2] )
Therefore, F3 = -3.9 N [ (cos 76 - cos 35) i + (sin35 - sin 76) j ]
= - 3.9 N [ -0.57723 i - 0.39672 j ]
F3 = 2.25120 N i + 1.5472 N j
You can see that this force is pulling up and to the right with slope m = 1.5472 / 2.2512 = 0.68728.
The reference angle is arctan (0.68728) = 34.5 degrees (above the positive x axis)
The magnitude is sqrt [ 1.5472 ^2 + 2.2512 ^2 ] = sqrt [7.4617] = 2.7316 N *
note: you would probably round the magnitude to 2.7 N if you just wanted 2 sig digits because that is the level of precision in the "given" information.
* It is probably helpful to sketch the triangle ABC, with A at the origin, with the sides AB and BC corresponding to the (magnitudes/direction of) forces F1 and F2 respectively.
The acceleration of object is zero means that the total sum of forces are zero. In other words it is equal and opposite force to combined the combined forces of F1 and F2.
I assume o to the right of a number means a degree.
Given
F1 = 3.9N at 76° below +x-axis
F2 = 3.9N at 35° above -x-axis
Decomposing the forces into x and y directions
F1x = 3.9*cos(-76) = 0.9435
F1y = 3.9*sin(-76) = -3.7841
F2x = 3.9*cos(180-35) = -3.1947
F2y = 3.9*sin(180-35) = 2.2369
Total of two component forces are
F12x = 0.9435 - 3.1947 = -2.2512
F12y = -3.7841 + 2.2369 = -1.5472
The third force is equal and opposite to these, thus
F3x = 2.2512
F3y = 1.5472
Adding the forces
F3 = √(2.2512^2 + 1.5472^2) = 2.7316
The angle is
Θ3 = arctan(1.5472/2.2512) = 34.4999
The magnitude of F3 is 2.7 N and the angle is 35° above +x-axis.