Angle of Alex's Force in Horizontal Tug of War | Find Answer Here

[RadiationX]

in a horizontal tug of war ,Alex,Betty, and Charles pull horizontally on an automobile tire at the angles shown in the overhead view (see attaced file) the tire remains stationary in spite of the three pulls alex pulls with a force of magnitude 220 Newtons and charles pulls with a force of magnitude of 170 Newtons the direction of charles' force is not given . what is the magnitude of bettys force?


what i need to know is when they find the components of Alex why do they use (180-47)=133 instead of the "natural choice of 47" ?

why do they use the 133 degree algle alexes' force makes with the positive x-axis insted of 47 degree algle this force makes with the negative x-axis.

choosing the wrong angle is totally messing me up big time. i can do the problem i just keep using the wrong angle. is there some general rule of what angle to use? my professor is terrible and my entire class is basically on it's own. i really need some understanding. thanks in adcance.

 

[gnome]

Can't see your diagram. What kind of file is that?

But, in general, since you have 3 vectors you want to add, each one has to be measured with reference to the same axis. You can use whatever axis you want but you have to use the same one for all of the vectors. Presumably it's less confusing to do it all in relation to the positive x-axis than to measure all the angles from the negative x-axis (but you should end up with the same answer either way, after you adjust your answer based on your choice of axis).

 

[Astronuc]

The angle is taken from the +x axis, so the angle is 133 degrees for Alex.

In a statics case, [itex]\Sigma F_x[/itex] = 0, and [itex]\Sigma F_y[/itex] = 0, and in order to add vector components, the angle must be taken from the same reference.

However one could write F(Alex) cos (47°) = F(Charles) cos ([itex]\theta_c[/itex]) where [itex]\theta_c[/itex] is the angle between x-axis and Charles's force vector.

 

[RadiationX]

thanks for the response. btw this a .doc file i made with microsoft word.

 

[RadiationX]

what if the angles were quandrantal? for example they use -90 degrees instead of 270 degrees for force(betty).

 

[learningphysics]

-90 degrees is the same angle as 270 degrees.

For your original question, you can use 47 or 133... it just depends on how you use it.

If you're using 47 (the natural angle so to speak)... then remember that Falex cos (47) is the magnitude of the force, and the vector is in the -x direction.

I would have chosen 47 myself as that's what I'm used to (dealing with magnitude and then afterwards... direction). Then I'd say:

Fcharles cos(theta) - Falex cos(47) =0

Here, we are dealing with the magnitude, and then dealing with the sign... meaning... Falex cos (47) gives the magnitude of the force, then we have to add a "-" sign since it's in the -x direction.

If you take the angle from the +x axis, then
Fcharles cos(theta) + Falex cos(133) =0

Both give the same result as cos(133) = -cos(47)

By using all the vectors with the same reference (from the +x-axis), you don't have to think about magnitude and sign separately. The sum of all the x-components is 0 (in statics):

Fa cos(theta_a) + Fb cos(theta_b) +... =0

It really depends on what the professor expects and what is best for you. You should be able to get the right answer both ways.

 

[gnome]

What if?

You choose whatever coordinate system you want to use, as long as you're consistent with yourself.

And if the question asked for the direction of a force, your answer is equally correct (or incorrect) whether you give it as a positive angle from the positive x-axis or a negative angle from the negative y axis, as long as you unambiguously explain how you are measuring it (unless the question specified which axis you must use).

In this case it's easy: they're only asking for the magnitude of Betty's force.

 

[RadiationX]

thanks ppl. i appreciate it.