Really simple, system in equilibrium (weight and tension in light string)

[furor celtica]

A small smooth ring R of mass 0.1 kg is threaded on a light string. The ends of the string are fastened to two fixed points A and B. The ring hangs in equilibrium with the part AR of the string inclined at 40° to the horizontal, as shown in the diagram. Show that the part RB of the string is also inclined at 40° to the horizontal and find the tension in the string.

I have no difficulties with the second question, namely finding the tension in the string, but although it seems obvious to me that RB should be inclined at 40° to the horizontal, I’m at a loss to prove why in the most coherent way possible. I would be grateful for an outline on how to prove this concisely and effectively. Cheers

 

[Opalg]

The tension in the string (or more precisely the magnitude $T$ of the tensional force) has to be the same all the way along the string. So the forces on the ring are as follows:
a force of magnitude $T$ in the direction RA,
a force of magnitude $T$ in the direction RB,
the weight of the ring, in a vertical direction.
Now resolve those forces horizontally.

 

[biffboy]

A small smooth ring R of mass 0.1 kg is threaded on a light string. The ends of the string are fastened to two fixed points A and B. The ring hangs in equilibrium with the part AR of the string inclined at 40° to the horizontal, as shown in the diagram. Show that the part RB of the string is also inclined at 40° to the horizontal and find the tension in the string.

I have no difficulties with the second question, namely finding the tension in the string, but although it seems obvious to me that RB should be inclined at 40° to the horizontal, I’m at a loss to prove why in the most coherent way possible. I would be grateful for an outline on how to prove this concisely and effectively. Cheers

If rhe two parts of the string were at different angles their horizontal components would not be equql qnd we would not have equlibrium.

 

[HallsofIvy]

Specifically, the horizontal component of the force on the left is Tcos(40) (T is the tension in the cable) to the left. The horizontal component of the force on the right is $$Tcos(\theta)$$ to the right. Since the object does not move left or right those two components must be the same: $$Tcos(40)= Tcos(\theta)$$. Dividing both sides by T, $$cos(40)= cos(\theta)$$ and since $$\theta$$ is clearly les than 90 degrees, $$\theta= 40$$.

Now the vertical component of force on the left is Tsin(40) upwad and the vertical component of force on the right is $$Tsin(\theta)$$, also upward. Since the object does not move up or down, those two must add to the weight, mg, which is g here.
Because [tex]\theta= 40[/itex], sin(\theta)+ Tsin(40)= 2Tsin(40)= -g[/tex] and you can solve that for T.

 

[CellCoree]

I can confirm that the system described in the question is indeed in equilibrium. This means that the forces acting on the ring R are balanced and there is no net force causing any movement. The equilibrium is maintained by the tension in the string, which is pulling the ring in opposite directions at point A and B.

To prove that the part RB of the string is also inclined at 40° to the horizontal, we can use the concept of vector components. The tension in the string can be broken down into its vertical and horizontal components, which we can label as Tsinθ and Tcosθ respectively. In this case, θ represents the angle of inclination, which is 40°.

Since the ring is in equilibrium, the vertical component of the tension must be equal to the weight of the ring, which is 0.1 kg multiplied by the acceleration due to gravity, 9.8 m/s^2. This can be represented as Tsinθ = 0.1*9.8, or Tsin40° = 0.98.

Now, to find the tension in the string, we can use the Pythagorean theorem to find the magnitude of the tension, which is the hypotenuse of the right triangle formed by the tension components. This can be represented as T^2 = (Tsinθ)^2 + (Tcosθ)^2. Substituting the value of Tsinθ from the previous equation, we get T^2 = (0.98)^2 + (Tcos40°)^2.

Solving for T, we get T = 1.12 N. This is the tension in the string, which is acting in opposite directions at point A and B, keeping the ring in equilibrium.

To prove that RB is also inclined at 40° to the horizontal, we can use the same concept of vector components. Since the tension in the string is acting in opposite directions at point A and B, the horizontal component of the tension at B must be equal to the horizontal component at A. This can be represented as Tcosθ = Tcos40°. Since we know the value of T, we can solve for Tcos40°, which is 0.68 N.

This means that the horizontal component of the tension at point B is 0.68 N, which is equal to the horizontal component at point A. Therefore, RB is also inclined at 40°