# Math Help - Equilibrium of particles

1. ## Equilibrium of particles

A block B of mass 0.4 kg and a particle P of mass 0.3 kg are connected by a light inextensible string. The string passes over a smooth pulley at the edge of a rough horizontal table. B is in contact with the table and the part of the string between B and the pulley is horizontal. P hangs freely below the pulley (see diagram).

Code:
(i) The systemis in limiting equilibrium with the string taut and P on the point of moving downwards.

Find the coefficient of friction between B and the table. [5]
Code:
(ii) A horizontal force of magnitude X N, acting directly away from the pulley, is now applied to B.

The system is again in limiting equilibrium with the string taut, and with P now on the point of moving upwards. Find the value of X.
I've done part i)

3 (i) T = 0.3g
F = T
R = 0.4g
F = μR
Coefficient is 0.75

but how do i do part 2?

I got W=T for P
so 0.3g=T

For B: T-F-X=0
0.3g-0.3g-X=0

wouldnt X=0?

X = 0.3g + 0.3g
X = 5.88N

2. net force on B ...

$0 = X - (f + T)$

net force on P ...

$0 = T - mg$

combine the equations ...

$0 = X - f - mg$

$X = f + mg$

$X = \mu Mg + mg = 0.3g + 0.3g = 0.6g$

3. Originally Posted by BabyMilo
For B: T-F-X=0
0.3g-0.3g-X=0
everything is correct up to here.

why are you saying x+F=T?
friction acts AGAINST motion, the question told you the system is about to move away from the pulley so friction force should be in the same direction as T.
so in fact X=T+F is the correct equation.

Hope that helps

4. Originally Posted by Krahl
everything is correct up to here.

why are you saying x+F=T?
friction acts AGAINST motion, the question told you the system is about to move away from the pulley so friction force should be in the same direction as T.
so in fact X=T+F is the correct equation.

Hope that helps
initally, i thought T would have some friction force to it as well.
never thought X was greater than T.

5. yep X is greater than T also there would be no motion if we apply a force X
$T-F \leq X \leq T+F$