Finding Equilibrium Forces: Are My Vector Calculations Correct?

[IGeekbot]

Ok, I am in an intro to physics, so it is all trig based. But I just need to know if I did these vector problems right.


an object in equilibrium has 3 forces exerted on it. one at 90 degress is 33N, one at 60 degrees is 44N. Find the third.


Using the sine and cosine of those vectors, i got the thrid force to be 74.3N at 253 degrees. Is that right?


ANd the other one is: five forces act on an object, the first is 60N at 90 degrees, the second is 40N at o degrees, the third is 80N at 270 degrees, te fourth is 50N at 60 degrees, and the fifth is 40N at 180 degrees. what's the sixth force that puts the object in equilibrium?

I know that the 2 40N forces cancel, so i just found the vertical and horizontal force of the other three forces. Then after using more sin and cos laws, and

Fnet6^2 =Fnetx^2 + Fnety^2

I got the force should be 33.97N, and using tanget, it should be at 43 degrees.

 

[Galileo]

Hi IGeekBot,

The first answer is correct. (Actually 74.4 N)

For the second, it should be at 223 degrees (180+43 degrees), watch the signs.
Since the sum of all forces is zero (equilibrium), you have:
$$F_6^2+F_{netx}^2+F_{nety}^2=0$$.
($F_{netx}$ and $F_{nety}$ do not include $F_6$)
The magnitude is 34.17 N. Be careful not to round off in intermediary stages of the calculation.

 

[Leong]

The first one is correct but for the seond one, i got 34.2 N at 223 degree from the positive x axis.

 

[IGeekbot]

thanks, our teacher is a bit hazy sometimes, and it sems th ewhole group was lost on this.