Question

Three ropes are tied in a knot as shown in the figure. One student pulls on rope A with 1.0 pound of force, and another student pulls on rope B with 7.0 pounds of force. How hard and in what direction must you pull on rope C to balance the first two pulls? Give the direction by specifying the angle (clockwise or counterclockwise) of the pull with the direction of rope A. 1. Rope B

Answer 1

Please do not downvote if angle taken by me are different, Since figure is not given, So please comment the value of angle and I will recalculate the final values.

Given that Net force at the knot is zero

Fnet = F1 + F2 + F3 = 0

F3 = -(F1 + F2)

Now Using given information:

Suppose direction of force on rope A is in x-axis

Fa = 1 lbf in x-axis

Fa = 1 i

Direction of force on rope B is 120 deg (I'm assuming this is 120 deg from cropped figure, if it's different let me know in the comment section)

Fb = 7 lbf relative to 120 deg from x-axis

Fb = Fbx i + Fby j

Fb = 7*cos 120 deg i + 7*sin 120 deg j

Fb = -3.5 i + 6.06 j

So third force on rope C will be

Fc = -(Fa + Fb)

Fc = -(1 i + (-3.5 i + 6.06 j))

Fc = (3.5 - 1) i - 6.06 j

Fc = -2.5 i - 6.06 j

from above we can see that Fc is in 3rd quadrant

Magnitude of third force will be

|Fc| = sqrt ((-2.5)^2 + (-6.06)^2)

|Fc| = 6.55 pound of force

Direction of Fc = arctan (Fcy/Fcx)

Direction = arctan (6.06/2.5) = 67.6 deg below -ve x-axis

Now since we need direction w.r.t to F1 OR +ve x-axis and F3 is in 3rd quadrant, So

Direction = 180 + 67.6 = 247.6 deg relative to the first force on rope A (Counter Clockwise)

Please Upvote.

Please Ask 2nd question as a new question, and attach figure of 2nd question too, I will be happy to help.