A lamp has a mass of 10kg and is supported by two wires that make an angle of 35 degrees between them. Find the tension in each wire?
6.
The tension is 6.
To find the tension in each wire, we can use the equilibrium conditions of the lamp. When the lamp is in equilibrium, the net force in both the horizontal and vertical directions is zero.
Let's start by considering the vertical equilibrium. The weight of the lamp acts vertically downward, and this force must be balanced by the vertical components of the tensions in the wires. Since the weight acts vertically downwards, the vertical component of tension in one wire must be equal to the weight:
Tension1 * sin(35°) = mg
where Tension1 is the tension in the first wire, m is the mass of the lamp, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, let's consider the horizontal equilibrium. The horizontal components of the tensions in the wires must balance each other out as there is no horizontal acceleration. The horizontal component of tension in one wire cancels out the horizontal component in the other wire as they act in opposite directions. Therefore, the horizontal components of the tensions are equal:
Tension1 * cos(35°) = Tension2 * cos(35°)
Simplifying this equation, we get:
Tension1 = Tension2
Now, we can substitute the value of Tension1 in terms of Tension2 from the vertical equilibrium equation:
Tension2 * sin(35°) = mg
Finally, we can solve for the tension in each wire:
Tension2 = mg / sin(35°)
Substituting the value of mass (10 kg) and sin(35°) into the equation, we can calculate the tension in each wire.