The three ropes in the figure are tied to a small, very light ring. Two of these ropes are anchored to walls at right angles with tensions T1 = 20 N and T2 = 60 N. What are the magnitude and direction of the tension vector T3 in the third rope?

Correct: Your answer is correct. N. ° (below the positive x-axis)

Let t be the angle made by T3 with the x-axis.

T3 cos(t) = 20 N ---- (1)
T3 sin(t) = 60 N ----- (2)

Square both equations and add:

T3^2 = 20^2 + 60^2 = 400 + 3600 = 4000
T3 = sqrt(4000) = 63.25 N

divide (2) by (1):
tan(t) = 60/20 = 3
t = arctan(3) = 71.57 degrees.

Since the angle clockwise from the x-axis some books may call it a negative angle and say the angle is -71.57 degrees or round it to -71.6 or even -72 degrees.

AFOKE88 AFOKE88 · Answer 2 · 2021-10-27T11:05:17-04:00

The magnitude and direction of the tension vector T3 in the third rope is;

63.25 N directed 71.565° below the horizontal.

We are given the tensions;

T1 = 20 N

T2 = 60 N

I have attached an image showing the rope with the 3 tensions showing T1, T2 and T3.

Since we are told that the ring is very light, we can ignore its weight.

Let us resolve the forces in the horizontal x-direction. Thus, we will have;

T1 = T3 cos θ - - - (eq 1)

Let us now resolve the forces in vertical y-direction to get;

T2 = T3 sin θ - - - (eq 2)

Let's divide eq(2) by eq(1) to get;

T2/T1 = T3 sin θ/T3 cos θ

T2/T1 = tan θ

Plugging in the relevant values gives;

60/20 = tan θ

θ = tan^(-1) 3

θ = 71.565°

Thus, from T1 = T3 cos θ , we have;

T3 = 20/cos 71.565

T3 = 63.25 N

Thus, the magnitude and direction of the tension vector T3 in the third rope is 63.25 N directed 71.565° below the horizontal.

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