Final answer:
The tension experienced by each of the two ropes suspending the 10 kg box at a 45-degree angle is approximately 69.3 N. The direction of the tension force is along the ropes, extending from the box to their attachment points on the ceiling.
Step-by-step explanation:
The student is asking about the tension in the ropes that are suspending a 10 kg box with both ropes at a 45-degree angle from the ceiling, which is a Physics problem involving static equilibrium and vector resolution. To find the tension in each rope, we need to consider the box is in static equilibrium, which implies the net force on the box is zero. Weight of the box acts downward, which is 10 kg × 9.8 m/s² (acceleration due to gravity, g), giving us 98 N.
Since the box is stationary and in equilibrium, the upward components of the tension in the two ropes must equal the weight of the box. Since the ropes are symmetrical about the box, the tension will be the same in each rope. Using trigonometry, specifically sine, because we are looking at the opposite side of the angle (the vertical component of the force), we get:
T × sin(45°) = 49 N
Therefore:
T = 49 N / sin(45°) ≈ 69.3 N
The direction of the force of tension will be along the ropes from the box towards the point where each rope is attached to the ceiling. If we considered the angles with respect to the box, they would both be 45 degrees to the horizontal.