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