A 46−N lithograph is supported by two wires. One wire makes a 28° angle with the vertical and the other makes a 16° angle with the vertical. Find the tension in each wire.
T1 cos 28 = T2 cos 16
(no net horizontal force)
T1 + T2 = 46
for vertical equilibrium
28o W. of N. = 118o CCW from +x-axis.
16o E. of N. = 74o CCW from +x-axis.
T1*Cos118 = -T2*Cos74.
T1 = -T2*Cos74/Cos118 = 0.587T2.
T1*sin118 + T2*sin74 = -(-46),
0.883T1 + 0.961T2 = 46,
0.883(0.587T2) + 0.961T2 = 46,
0.518T2 + 0.961T2 = 46, T2 = 31.1 N.
T1 = 0.587*T2 = 0.587*31.1 = 18.2 N.
To find the tension in each wire, we can divide the lithograph's weight between the two wires according to their angles.
1. First, let's consider the wire with a 28° angle. We can find the vertical component of the tension by using the sine function:
T1 (vertical component) = T1 * sin(28°)
2. Similarly, for the wire with a 16° angle, we can find the vertical component of the tension:
T2 (vertical component) = T2 * sin(16°)
3. Since the lithograph is in equilibrium, the vertical components of the tensions must add up to balance out the weight of the lithograph:
T1 (vertical component) + T2 (vertical component) = weight of lithograph
4. The weight of the lithograph is given as 46 N.
5. Now we can substitute the known values into the equation:
T1 * sin(28°) + T2 * sin(16°) = 46 N
6. We also know that the total tension in both wires must equal the weight of the lithograph. Therefore,
T1 + T2 = 46 N
7. We have two equations and two unknowns (T1 and T2), so we can solve the system of equations.
Based on the information provided, we can calculate the tension in each wire.
To find the tension in each wire, we can use the concept of resolving forces.
Let's call the tension in the wire that makes a 28° angle with the vertical "T1" and the tension in the wire that makes a 16° angle with the vertical "T2".
We can start by resolving the forces vertically. Since the lithograph is in equilibrium, the vertical component of the tension in both wires must balance the weight of the lithograph.
The vertical component of T1 can be calculated as T1 * cos(28°), and the vertical component of T2 can be calculated as T2 * cos(16°).
Setting up the equation, we have:
T1 * cos(28°) + T2 * cos(16°) = 46 N
Next, we can resolve the forces horizontally. The horizontal components of T1 and T2 must balance each other since the lithograph is not moving horizontally.
The horizontal component of T1 can be calculated as T1 * sin(28°), and the horizontal component of T2 can be calculated as T2 * sin(16°).
Setting up the equation, we have:
T1 * sin(28°) = T2 * sin(16°)
Now we have a system of two equations with two variables (T1 and T2).
We can solve these equations simultaneously to find the values of T1 and T2.
Using substitution or elimination method, we can solve for T1:
From the second equation, we have:
T1 = (T2 * sin(16°)) / sin(28°)
Substituting this value of T1 into the first equation, we have:
(T2 * sin(16°)) / sin(28°) * cos(28°) + T2 * cos(16°) = 46 N
Now we can solve for T2:
(T2 * sin(16°) * cos(28°)) / sin(28°) + T2 * cos(16°) = 46 N
T2 * (sin(16°) * cos(28°) / sin(28°) + cos(16°)) = 46 N
T2 * (0.358 + 0.960) = 46 N
T2 * 1.318 = 46 N
T2 = 46 N / 1.318
T2 ≈ 34.87 N
Now that we have the value of T2, we can substitute it back into the second equation to solve for T1:
T1 = (T2 * sin(16°)) / sin(28°)
T1 = (34.87 N * sin(16°)) / sin(28°)
T1 ≈ 19.75 N
Therefore, the tension in the wire that makes a 28° angle with the vertical (T1) is approximately 19.75 N, and the tension in the wire that makes a 16° angle with the vertical (T2) is approximately 34.87 N.