Tension? knowing friction. I have the answer but need explaination

[Giiang]

Tension? knowing friction. I have the answer but need explanation :(

Homework Statement

A ring of mass 4 kg is attached to one end of a light string. The ring is threaded on a fixed horizontal rod and the string is pulled at an angle of 25◦ below the horizontal (see diagram). With a tension in the string of T N the ring is in equilibrium. (first image)

(i) Find, in terms of T, the horizontal and vertical components of the force exerted on the ring by
the rod.

The coefficient of friction between the ring and the rod is 0.4.

(ii) Given that the equilibrium is limiting, find the value of T.

=> Summary:
m = 4kg
angle = 25◦
μ=0.4

Homework Equations

F=μR

The Attempt at a Solution

(second image)

(i) R (→) Tcos25
R (up) Tsin25 + 40
(ii) The answer says Tcos25 = μR + 0.196T
Why? What is 0.196? My first thought was Tcos25 = μR but it appears to be wrong. Help please D:

 

[Andrew Mason]

Homework Statement

A ring of mass 4 kg is attached to one end of a light string. The ring is threaded on a fixed horizontal rod and the string is pulled at an angle of 25◦ below the horizontal (see diagram). With a tension in the string of T N the ring is in equilibrium. (first image)

(i) Find, in terms of T, the horizontal and vertical components of the force exerted on the ring by
the rod.

The coefficient of friction between the ring and the rod is 0.4.

(ii) Given that the equilibrium is limiting, find the value of T.

=> Summary:
m = 4kg
angle = 25◦
μ=0.4

Homework Equations

F=μR

The Attempt at a Solution

(second image)

(i) R (→) Tcos25
R (up) Tsin25 + 40
(ii) The answer says Tcos25 = μR + 0.196T
Why? What is 0.196? My first thought was Tcos25 = μR but it appears to be wrong. Help please D:

Welcome to PF Gilang!

It is .169T which is the additional friction force due to the vertical component of the string tension: μTsin25 = .4T(.423) = .169T

AM

 

[Giiang]

Thank you. However, could you please elaborate? I'm still quite confused. Tcos25 is the horizontal component, why does it equal the vertical component?

 

[Andrew Mason]

Thank you. However, could you please elaborate? I'm still quite confused. Tcos25 is the horizontal component, why does it equal the vertical component?

It doesn't.

The maximum force of static friction (ie. the maximum horizontal force that the string can apply) is equal to the normal force on the rod multiplied by μ_s, the co-efficient of static friction. That normal force is: mg + Tsinθ. So F_max-horiz= μ_s(mg + Tsinθ)

So, the maximum tension is the tension whose horizontal component meets, but does not exceed, the maximum force of static friction. That is where your Tcosθ comes in.

AM

 

[HalfLostAndSearching]

Tension and friction are two important forces that play a role in the equilibrium of the ring in this scenario. Tension is the force that is exerted on the ring by the string, while friction is the force that opposes the motion of the ring along the horizontal rod. In order to fully understand the equilibrium of the ring, it is important to consider both of these forces.

In the first part of the solution, the horizontal and vertical components of the force exerted on the ring by the rod are calculated using the trigonometric relationships between the angles and sides of the triangle formed by the string and the rod. The horizontal component is equal to Tcos25, while the vertical component is equal to Tsin25 + 40.

In the second part of the solution, the equation F=μR is used to calculate the force of friction, where μ is the coefficient of friction and R is the normal force exerted on the ring by the rod. Since the equilibrium is limiting, the force of friction must be equal to the maximum force of static friction, which is given by μR. This means that Tcos25 must be equal to μR, and since μ=0.4, we can substitute this into the equation to get Tcos25=0.4R.

However, in the given solution, an additional term of 0.196T is added to the equation. This is because the force of friction is not the only force acting on the ring in the horizontal direction. There is also a component of the tension force, Tsin25, that is acting against the force of friction. This component must be subtracted from the force of friction to get the net force acting on the ring in the horizontal direction, which is equal to Tcos25. Therefore, the equation becomes Tcos25=μR-Tsin25. By substituting in the values for μ and R, we get Tcos25=0.4(Tsin25+40)-Tsin25. Simplifying this equation results in Tcos25=0.196T, which is the same as the given solution.

In conclusion, the additional term of 0.196T is necessary to account for the component of the tension force that is acting against the force of friction. This is important in determining the value of T, which is necessary for the equilibrium of the ring. I hope this explanation helps to clarify the solution for you.