A 200 N mass is supported by two wires that make angles of 54 degrees and 67 degrees with the ceiling.
Determine the tension in each wire.
Make 2 diagrams, one called the position diagram, which shows the
actual situation, marking the angles at the ceiling and showing the hanging
mass of 200 N.
the other is a vector diagram where the sides of the triangle show the tension
I labelled mine ABC where angle A = 23°, angle B = 121° and angle C = 36°
with AC = 200
by the sine law:
AB/sin36 = 200/sin121
AB = 200sin36/sin121 = 137.146 N
BC/sin23 = 200/sin121
BC= 200sin23/sin121 = 91.168 N
check my arithmetic
All angles are measured CCW from +x-axis.
T1*sin36+T2*sin157 = 200
Eq1: 0.59T1 + 0.39T2 = 200.
T1*cos36 = -(T2*cos157)
0.81T1 = -(-.92T2)
T1 = 1.14T2.
In Eq1, replace T1 with 1.14T2 and solve for T2:
0.59*1.14T2 + 0.39T2 = 200
0.67T2 + 0.39T2 = 200
1.06T2 = 200
T2 = 188.7 N.
T1 = 1.14*188.7 = 215 N.
To determine the tension in each wire, we can use the concept of resolving forces. The weight of the mass (200 N) can be thought of as a force acting vertically downwards. Let's call this force W.
We have two wires supporting the mass, making angles of 54 degrees and 67 degrees with the ceiling. Let's call the tensions in the first and second wires T1 and T2, respectively.
To find the tension in each wire, we need to resolve the weight force (W) into its components along the wires.
The vertical component of the weight force (Wv) can be found using the equation:
Wv = W * sin(angle)
The horizontal component of the weight force (Wh) can be found using the equation:
Wh = W * cos(angle)
Now, let's calculate the vertical and horizontal components of the weight force:
For the first wire with an angle of 54 degrees:
Wv1 = W * sin(54)
Wh1 = W * cos(54)
For the second wire with an angle of 67 degrees:
Wv2 = W * sin(67)
Wh2 = W * cos(67)
Since the sum of the vertical forces must balance the weight (no vertical acceleration), we have:
T1 * cos(54) + T2 * cos(67) = W
And since the sum of the horizontal forces must be zero (no horizontal acceleration), we have:
T1 * sin(54) - T2 * sin(67) = 0
We can now solve these two equations simultaneously to find the values of T1 and T2.
1. Solve the first equation for T2:
T2 = (W - T1 * cos(54)) / cos(67)
2. Substitute this value of T2 into the second equation:
T1 * sin(54) - (W - T1 * cos(54)) / cos(67) * sin(67) = 0
Now, we can solve this equation to find the value of T1:
T1 * sin(54) - (W - T1 * cos(54)) / cos(67) * sin(67) = 0
Simplify the equation:
T1 * sin(54) - (W * sin(67) - T1 * cos(54) * sin(67)) / cos(67) = 0
Multiply through by cos(67):
T1 * sin(54) * cos(67) - W * sin(67) + T1 * cos(54) * sin(67) = 0
Rearrange the equation:
T1 * (sin(54) * cos(67) + cos(54) * sin(67)) = W * sin(67)
Use the trigonometric identity sin(a + b) = sin(a) * cos(b) + cos(a) * sin(b):
T1 * sin(54 + 67) = W * sin(67)
T1 * sin(121) = W * sin(67)
Substitute the value of W = 200 N:
T1 * sin(121) = 200 * sin(67)
Calculate sin(121) and sin(67) using a scientific calculator:
T1 * 0.8776 = 200 * 0.9211
Now, solve for T1:
T1 = (200 * 0.9211) / 0.8776
Calculate T1:
T1 ≈ 210.15 N
Finally, substitute this value of T1 back into the first equation to find T2:
T2 = (W - T1 * cos(54)) / cos(67)
T2 = (200 - 210.15 * cos(54)) / cos(67)
Calculate T2:
T2 ≈ 145.68 N
Therefore, the tension in the first wire is approximately 210.15 N, and the tension in the second wire is approximately 145.68 N.