A box weighing 415 N is hanging from two chains attached to an overhead bean at angles of 56 degrees and 49 degree. Find the magnitude of the tension in each chain algebraically.
After sketching a "position diagram", that is, a diagram showing the actual situation,
I made a vector diagram with a vertical side labelled 415 N, the side opposite the angle of 41°
as T1, and the side opposite the 34° angle as T2. The 3rd angle would be 105°
by the sine law:
T1/sin41 = 415/sin105
T1 = 415sin41/sin105 = ...
find T2 in the same way
Did you find the answer?
To find the magnitude of the tension in each chain algebraically, we can use trigonometric principles.
Let's label the tensions in the two chains as T1 and T2. We can break down the weight of the box into vertical and horizontal components.
The vertical component of the weight is given by W * cos θ, where W is the weight and θ is the angle between the weight vector and the vertical axis. In this case, θ = 56 degrees and 49 degrees. So the vertical components of the weight are:
W1_vertical = W * cos θ1 = 415 N * cos 56°
W2_vertical = W * cos θ2 = 415 N * cos 49°
Since the tension in each chain balances out the vertical component of the weight, we have:
T1 = W1_vertical
T2 = W2_vertical
Substituting the given values into the equations, we find:
T1 = 415 N * cos 56° ≈ 230.96 N
T2 = 415 N * cos 49° ≈ 279.48 N
Therefore, the magnitude of the tension in each chain, algebraically, is approximately 231 N for T1 and 279 N for T2.