A picture hangs motionless on a wall as shown in the Figure. If the picture has a mass of 5.0 kg, what are the tension forces in the wires?
You're going to somehow show the figure before anyone can answer your question.
To determine the tension forces in the wires, we need to consider the forces acting on the picture hanging motionless on the wall. In this case, there are two wires attached to the picture that are pulling on it, providing the upward tension forces.
To find the tension forces, we can start by analyzing the vertical equilibrium of forces. Since the picture is motionless, it means that the net force acting on it in the vertical direction is zero.
We can break down the picture's weight into two components: one pulling downwards parallel to the wall, and the other perpendicular to it. The component parallel to the wall is balanced by the static friction between the picture and the wall, preventing it from sliding down.
The downward force component perpendicular to the wall can be balanced by the sum of the upward tension forces exerted by the wires. Since there are two wires, the tension forces will be equal.
Therefore, we can set up the equation:
2T = mg,
where T is the tension force in each wire, m is the mass of the picture (5.0 kg), and g is the acceleration due to gravity (9.8 m/s^2).
Solving for T, we have:
T = (mg) / 2.
Plugging in the given values:
T = (5.0 kg) * (9.8 m/s^2) / 2.
Calculating this expression gives us the tension force in each wire.