A500 man stands at the center of the rope such that each half of the ropes makes an angle of 10 with the horizentle . What is the tension of the rope ???
mg=500 N
vector T + vecrorT +vector mg = 0
y-projection : 2Tsinα = mg
T=mg/2sinα=500/2sin10=1439.7 N
To find the tension of the rope, we can use the concept of equilibrium in a system. When an object is in equilibrium, the net force acting on it is zero. In this case, we can consider the forces acting on the person standing at the center of the rope.
Since each half of the rope makes an angle of 10° with the horizontal, the total angle between the two halves is 20°. We can break down the tension force in each half of the rope into horizontal and vertical components.
Let's assume the tension force in each half of the rope is T. Since the person is standing in the center, the vertical components of each half of the rope's tension force cancel each other out.
The horizontal components of the tension forces must balance the weight of the person, which is given as 500 N. The horizontal component is equal to T * cos(20°), and since there are two halves, we get:
2 * T * cos(20°) = 500 N
Now, we can solve for T:
T = 500 N / (2 * cos(20°))
Using a calculator, we can find:
T ≈ 500 N / (2 * 0.9397)
T ≈ 267.97 N
Therefore, the tension of the rope is approximately 267.97 N.