najinaji
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#1
Report Thread starter 8 years ago
#1
Currently going over a past paper question:

'Two particles A and B, of mass m and 2m respectively, are attached to the ends of a light inextensible string. The particle A lies on a rough horizontal table. The string passes over a small smooth pulley P fixed on the edge of the table. The particle B hangs freely below the pulley. The coefficient of friction between A and the table is mu. The particles are released from rest with the string taut. Immediately after release, the magnitude of the acceleration of A and B is 4/9g. By writing down separate equations of motion for A and B'

(a) find the tension in the string immediately after the particles begin to move,

(b) show that mu = 2/3

When B has fallen a distance h, it hits the ground and does not rebound. Particle A is then a distance 1/3 h from P.
(c) Find the speed of A as it reaches P.

b) Uses the formula T-mu(g)=ma and c) uses the formula ma=mu(mg) Where did these formulas come from? :confused:
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raheem94
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#2
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(Original post by najinaji)
Currently going over a past paper question:

'Two particles A and B, of mass m and 2m respectively, are attached to the ends of a light inextensible string. The particle A lies on a rough horizontal table. The string passes over a small smooth pulley P fixed on the edge of the table. The particle B hangs freely below the pulley. The coefficient of friction between A and the table is mu. The particles are released from rest with the string taut. Immediately after release, the magnitude of the acceleration of A and B is 4/9g. By writing down separate equations of motion for A and B'

(a) find the tension in the string immediately after the particles begin to move,

(b) show that mu = 2/3

When B has fallen a distance h, it hits the ground and does not rebound. Particle A is then a distance 1/3 h from P.
(c) Find the speed of A as it reaches P.

b) Uses the formula T-mu(g)=ma and c) uses the formula ma=mu(mg) Where did these formulas come from? :confused:
Resolving the forces on A gives the equation,  T - \mu R = ma , we know R=mg.
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Phredd
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Both equations you have asked about are variants on F=ma, where F is the resultant force. In the first case, the resultant force is the tension minus the friction, and in the second case the resultant force is just the friction.

for example, the first equation:


F=ma

\ \ \ \ \ \ \ \ \ \ F = T - F_r

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ F_r = \mu R

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ R = mg

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ F_r = \mu mg

T - \mu mg = ma
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najinaji
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(Original post by raheem94)
Resolving the forces on A gives the equation,  T - \mu R = ma , we know R=mg.
Ah, I see. But why is there a mu there?

(Original post by Phredd)
Both equations you have asked about are variants on F=ma, where F is the resultant force. In the first case, the resultant force is the tension minus the friction, and in the second case the resultant force is just the friction.

for example, the first equation:


F=ma

\ \ \ \ \ \ \ \ \ \ F = T - F_r

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ F_r = \mu R

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ R = mg

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ F_r = \mu mg

T - \mu mg = ma
...I hate Mechanics.
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raheem94
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(Original post by najinaji)
Ah, I see. But why is there a mu there?


...I hate Mechanics.
A is on the rough horizontal table, hence friction acts on it. So mu is there.

I like mechanics the most, it is really interesting if you understand it.
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Phredd
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(Original post by najinaji)
Ah, I see. But why is there a mu there?
The  \mu is the coefficient of friction, by definition  F_r = \mu R
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Niamhkelly533
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What year of past paper is this??
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