Why taking angle 110 instead of 70 for finding the magnitude of F2 in X and Y components?

[Benjamin_harsh]

Homework Statement

why taking angle 110 instead of 70 for finding the magnitude of F2 in X and Y components?

Relevant Equations

F2x = 150.cos 110 = - 51.30, F2y = 150.sin 110 = 140.95

F1x = 120. Cos 300 = 103.92 ; F1y = 120.sin 300 = 60N
F2x = 150.cos 110 = - 51.30, F2y = 150.sin 110 = 140.95

why taking angle 110 instead of 70 for finding the magnitude of F2 in X and Y components?

 

[lomidrevo]

If you take angle ##70°## insteady of ##110°##, do you get negative value for ##F_{2x}## component (as it should be according to the diagram)?

 

[kuruman]

If you are interested in just the magnitude of ##F_{2}## it doesn't matter which angle you take. That's because
##F_2=\sqrt{F_2^2\sin^2(\theta)+F_2^2\cos^2(\theta)}.## Note that ##\cos(110^o)=-\cos(70^o)## and when you square, the negative sign drops out.

 

[collinsmark]

I can't let this go. The diagram is very misleading.

The diagram shows the angular difference between [itex] F_1 [/itex] and [itex] F_2 [/itex] as [itex] 110^o [/itex]. But that doesn't make any sense. That would make the total angle from the positive x-axis to the negative x-axis as [itex] 210^o [/itex]. But that's impossible, since the x-axis is a straight line, and all angles that span a straight line are [itex] 180^o [/itex]. So first and foremost, that needs to be fixed.

From here on out, I'm assuming that the [itex] 110^o [/itex] angle spans from the positive x-axis to [itex] F_2 [/itex] (not [itex] F_1 [/itex] to [itex] F_2 [/itex] as is it shown in the original post).

From there, just follow @kuruman's and @lomidrevo's advice. If you use the [itex] 70^o [/itex] angle, which is the angle with respect to the negative x-axis, then the [itex] \cos 70^o [/itex] will be projected along the negative x-axis. That's a tidbit you need to keep in the back of your mind when interpreting the result. However, if you use the [itex] 110^o [/itex] angle, which is the angle with respect to the positive x-axis, the [itex] \cos 110^o [/itex]result is already negative so it takes care of the negative sign automatically.

 

[kuruman]

From here on out, I'm assuming that the ##110^o## angle spans from the positive x-axis to ## F_2## (not ##F_1## to ##F_2## as is it shown in the original post).

That was my interpretation of the situation. My reading was that OP was questioning whether the difference between ##\cos(70^o)## and ##\cos(110^o)## is more than just an algebraic sign.