A sign weighing 10N hangs from two ropes. Each rope is 45 degrees from the horizontal.
1. What is the tension in each rope?
Let T be the rope tension force in each. They are the same in each rope, due to symmetry.
2 T cos45 = 10 N (vertical equilibrium)
T = 10N * sqrt2 = 14.14 N
T1*sin(180-45) + T2*sin45 = -10*sin270.
0.707T1 + 0.707T2 = 10.
T1 = T2.
0.707T1 + 0.707T1 = 10.
1.414T1 = 10.
T1 = 7.07 N. = T2.
To find the tension in each rope, we can use trigonometry and resolve the forces acting on the sign.
Let's assume that the tension in one rope is T1 and the tension in the other rope is T2.
1. Resolve the weight of the sign into its horizontal and vertical components:
The vertical component of the weight is given by:
vertical component = weight × cos(angle)
vertical component = 10N × cos(45°)
The horizontal component of the weight is given by:
horizontal component = weight × sin(angle)
horizontal component = 10N × sin(45°)
2. Since the sign is in equilibrium, the total vertical forces must balance each other. Therefore, the vertical component of the tensions in the two ropes must equal the vertical component of the weight.
T1 × cos(45°) + T2 × cos(45°) = 10N × cos(45°)
3. Similarly, the total horizontal forces must balance each other. Therefore, the horizontal component of the tensions in the two ropes must equal the horizontal component of the weight.
T1 × sin(45°) + T2 × sin(45°) = 10N × sin(45°)
4. We have two equations with T1 and T2 as unknowns. We can solve these equations simultaneously to find their values.
T1 × cos(45°) + T2 × cos(45°) = 10N × cos(45°)
T1 × sin(45°) + T2 × sin(45°) = 10N × sin(45°)
Dividing both equations by cos(45°) and sin(45°) respectively, we get:
T1 + T2 = 10N
T1 + T2 = 10N
Adding these two equations together, we get:
2T1 + 2T2 = 20N
Dividing by 2, we find:
T1 + T2 = 10N
Since we have two equations that are equal:
T1 + T2 = 10N
T1 + T2 = 10N
We can conclude that:
T1 = T2 = 10N / 2 = 5N
So, the tension in each rope is 5N.