# Homework Help: Tension? knowing friction. I have the answer but need explaination

1. Apr 20, 2014

### Giiang

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

1. The problem statement, all variables and given/known data

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

2. Relevant equations

F=μR

3. 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:

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Last edited: Apr 20, 2014
2. Apr 20, 2014

### Andrew Mason

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

3. Apr 21, 2014

### 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?

4. Apr 21, 2014

### Andrew Mason

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 μs, the co-efficient of static friction. That normal force is: mg + Tsinθ. So Fmax-horiz= μs(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

Last edited: Apr 21, 2014