A man weighing 500 newtons sits in swing suspended with two ropes if each rope makes an angle of 30 degree with the vertical, what is the tension in each rope?
sketch the diagram. First, assume only one rope.
vertical load on the hinge: mg
tension= mg/cos30
so tension on each of two ropes is half that. weight mg = 500
tensioneachrope=250/cos30
To find the tension in each rope, we need to use the principles of equilibrium.
First, let's draw a diagram to visualize the situation. We have a swing with a man weighing 500 newtons, and two ropes suspending the swing. Each rope makes an angle of 30 degrees with the vertical.
Let's assume that the tension in each rope is T (since both the ropes have the same tension).
Now, consider the forces acting on the swing:
1. The weight of the man (500 N) acts vertically downwards.
2. The tension in each rope acts along the direction of the ropes.
Since the swing is at rest (equilibrium), the sum of the forces in the vertical direction must be zero. This means that the vertical component of the tension in each rope must balance out the weight of the man.
The vertical component of the tension can be found using trigonometry. We know that the angle between the rope and the vertical is 30 degrees.
Using the formula:
Vertical component = T * cos(angle)
Vertical component = T * cos(30 degrees)
Now let's solve for the vertical component of the tension:
Vertical component = T * cos(30 degrees)
= T * (sqrt(3)/2)
Since the weight of the man is 500 N and it is balanced by the vertical component of the tension, we have:
T * (sqrt(3)/2) = 500 N
Solving for T:
T = (500 N) / (sqrt(3)/2)
= (500 N) * (2/sqrt(3))
= (1000/sqrt(3)) N
So, the tension in each rope is approximately (1000/sqrt(3)) N.