You pull at an angle of 30 ° with respect to the horizontal. One friend pulls with a force of 30 𝑁 to the left. Your other friend pulls with a force of 10 𝑁 downward. How much force must you apply to your rope (the tension in the rope) so the ring is in static equilibrium?
To find the amount of force you must apply to your rope (the tension in the rope) so the ring is in static equilibrium, we need to consider the forces acting on the ring and set up an equilibrium equation.
Let's break down the forces acting on the ring:
1. Your friend pulling with a force of 30 N to the left: This force acts horizontally and can be represented as F_{H} = -30 N (negative sign indicates leftward direction).
2. Your other friend pulling with a force of 10 N downward: This force acts vertically and can be represented as F_{V} = -10 N (negative sign indicates downward direction).
3. The force you apply to your rope: This force acts at an angle of 30° with respect to the horizontal. Let's represent this force as F_{T}.
The equilibrium condition states that the sum of all forces acting on the ring must be zero for it to be in static equilibrium. We can set up the following equation:
ΣF_{H} + ΣF_{V} + ΣF_{T} = 0
Substituting the known values:
-30 N + (-10 N) + F_{T} = 0
Simplifying the equation:
F_{T} = 40 N
Therefore, you must apply a force of 40 N to your rope (the tension in the rope) to keep the ring in static equilibrium.