A light inextensible string, which passes over a fixed smooth peg A, is
fastened at one end to a small ring R of mass 0.1kg and at the other end to a
particle P of mass 0.3kg. The ring is threaded on a fixed rough vertical wire
and the system rests in equilibrium with the part AR of the string inclined at
an angle of 60 degrees to the vertical and the part AP of the string vertical
as shown in the diagram.Draw a clear diagram showing all forces acting on the particle and on the ring and state the tension in the string.Given that the equilibrium is limiting, explain why the ring is on the point of moving upwards and calculate the coefficient of friction between the ring and
the wire.Find the magnitude of the resultant force exerted on the peg by the string.
To answer the question, let's start by drawing a diagram of the forces acting on the particle P and the ring R.
First, draw a vertical line to represent the fixed rough wire. Label it as the vertical axis.
Next, draw a straight line segment for the string. Label the left end as A (where it is attached to the smooth peg) and the right end as R (where the ring is attached). Label the middle point of the string as P (where the particle is located).
Now, draw arrows to represent the forces acting on the particle P and the ring R.
For the particle P:
- Draw an arrow pointing upward from P to represent the tension force in the string acting on the particle. Label this force as T.
- Draw an arrow pointing downward from P to represent the weight of the particle (mg, where m is the mass of the particle and g is the acceleration due to gravity).
For the ring R:
- Draw an arrow pointing downward from R to represent the weight of the ring (mg, where m is the mass of the ring and g is the acceleration due to gravity).
- Draw an arrow pointing upward from R to represent the tension force in the string acting on the ring. Label this force as T.
Now let's address the first part of the question - determining the tension in the string.
Since the system is in equilibrium, the net force on both the particle and the ring is zero. This means that the tension in the string is equal to the weight of the particle and the ring combined.
T = (mass of particle + mass of ring) * g
= (0.3kg + 0.1kg) * 9.8 m/s^2
= 0.4kg * 9.8 m/s^2
= 3.92 N
Now let's move on to explaining why the ring is on the point of moving upwards and calculating the coefficient of friction between the ring and the wire.
In order for the ring to be on the point of moving upwards, the friction force between the ring and the rough wire must be equal to the maximum possible static friction force. This means that the static friction force is at its limit and any additional force will cause the ring to start moving.
The equation for the maximum static friction force is given by:
Friction force (f) = coefficient of friction (μ) * normal force (N)
The normal force (N) is equal to the weight of the ring since it is in equilibrium:
N = mass of ring * g
= 0.1kg * 9.8 m/s^2
= 0.98 N
Now we can calculate the coefficient of friction by rearranging the equation and substituting the values:
f = μ * N
Since the ring is on the point of moving, the friction force is equal to the maximum static friction force:
f = μ * N = f_max
Therefore:
μ * N = f_max
Substituting the values:
μ * 0.98 N = f_max
We need to calculate the coefficient of friction (μ):
μ = f_max / N
= (μ * N) / N
= 1
Therefore, the coefficient of friction between the ring and the wire is 1.
Finally, let's find the magnitude of the resultant force exerted on the peg by the string.
The resultant force exerted on the peg is equal to the tension in the string, T, since there are no other external forces acting on the system.
Therefore, the magnitude of the resultant force exerted on the peg by the string is 3.92 N.
I hope this explanation helps! If you have any further questions, feel free to ask.