Question

In a physics laboratory class, three massless ropes are tied together at a point. A pulling force is applied along each rope:
F1 = 159 N at 60°,
F2 = 199 N at 100°,
F3 = 110 N at 190°.

a) What is the magnitude of a fourth force that acts to keep the point at the center of the system stationary?


b) What is the angle at which this fourth force acts to keep the point at the center of the system stationary? (Measure angle counterclockwise from the positivex-axis.

Answer 1

Start off with a force balance both for the x direction and the y - direction, adding the missing variables F and θ for our massless string andcounter-force:

ΣF_x = 0 = 159cos60 + 199cos100 + 110cos190 + Fcosθ = 0

ΣF_y = 0 = 159sin60 + 199sin100 + 110sin190 + Fsinθ = 0

Now solve for both Fcosθ and Fsinθ:

Fcosθ = -(159cos60 + 199cos100 + 110cos190)≈ 63.38 N

Fsinθ = -(159sin60 + 199sin100 + 110sin190)≈ -314.57 N

Now remember the trigonometric property:

F =√((Fcosθ)²+ (Fsinθ)²)

So we may then conclude that:

F =√(63.38²+ (-314.57)²) = 320.89 N

Finally we can solve forθ:

320.89cosθ = 63.38 N

cosθ = 63.38/320.89

θ = cos^-1(63.38/320.89) = 78.61^o

Please remember to check my numbers!

The process should be all there though!

Cheers,

Ian