Two ropes pull on a ring. One exerts a 47 N force at 30.0°, the other a 47 N force at 60.0°?

Question

Two ropes pull on a ring. One exerts a 47 N force at 30.0°, the other a 47 N force at 60.0°
(a) What is the net force on the ring?
Magnitude
____ N
Direction
____ °
(b) What are the magnitude and direction of the force that would cause the ring to be in equilibrium?
____ N
____ °

Answer 1

Break the forces into components. We'll use 0 degrees as the direction of the +x axis and 90 degrees as the direction of the +y axis.

Forces along the x axis:

Fx = 47 N cos 30 + 47 N cos 60

Forces along the y axis:

Fy = 47 N sin 30 + 47 N sin 60

Add each up to get Fx and Fy. Then find the net force using the Pythagorean Theorem: F = syqrt (Fx ^2 + Fy ^2 ). The direction is inverse-tangent Fy / Fx, measured from the +x axis.

Answer 2

remedy it graphically via drawing a line of fixed length at 25 levels to the vertical, say 300mm long to represent the linear stress of 600N. From one end of this line, draw a 2d line of the comparable length at an attitude of a hundred thirty levels to the 1st one. (one hundred eighty levels - 25 levels - 25 levels). degree the gap between the outer ends of the strains. each and each millimetre will characterize 2N. you additionally can do it trigonometrically using the accepted answer for a triangle the place 2 adjoining facets and all the angles are time-honored. this might characterize the entire stress resisting forward action, that would contain wind plenty. For the hydronamic resistive stress (drag) with the aid of water on my own, the wind load that are in any path must be time-honored and delivered into consideration.

Answer 3

Oh man! You beat me to the punch!

Like I really knew the answer. Yeah, right!