Mechanics - Friction

Kerrin · 2011-02-28 07:59:36

Two loads P and Q, of masses 2kg and 3kg respectively, are connected by a light inextensible string. Load P is on a rough horizontal table. The string passes over a smooth pulley fixed at the edge of the table, and load Q hangs vertically. A force of magnitude 30N acting at 30 degrees to the table is just sufficient to keep P in equilibrium. This force acts in the vertical plane containing the string.

Modelling the loads as particles, show that the coefficient of friction between P and the table is approximately 0.75.

The force of magnitude 30N is now removed. Ignoring air resistance, find the tension in the string in the ensuring motion.

Please help! Friction isn't one of my strong suits, and I'm having real trouble with this question!

byronjordan · 2011-02-28 15:41:27

A Newton of force accelerates 1 kg at 1 m/sec^2. The acceleration of gravity is 9.81 m/sec^2. So the force of gravity is 9.81 Newtons on 1kg of mass. Simply looking it up will confirm this.

Define down as vertical + and from P toward the pulley as horizontal +.
---------
Due to gravity pulling down on Q, it produces a horizontal force on P of +(3kg*9.81N/kg)=29.43N. (1)

Due to gravity, P has a vertical force down on itself of +(2kg*9.81N/kg)=19.62N. (2)

The 30N angled up force, produces
-(30N*cos(30°)) = -30N*0.866 = -25.98N horizontally (3)
and -(30N*sin(30°))=-30N*0.5=-15N vertically. (4)

So P has a horizontal force of
29.43N - 25.98N = 3.45N, From (1)&(3) : (5)
and a vertical force of 19.62N - 15N = 4.62N. From (2)&(4) : (6)

The coefficient of friction is the horizontal force divided by the vertical force which is
3.45N/4.62N = 0.747 ≈ 0.75. From (5)&(6) : (7)
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You can treat P's and Q's mass together and determine the acceleration of P and Q with the total forces acting on them. Then the force needed to accelerate just P plus the friction would be the tension in the string.

Once the 30N force is removed, the force down on P is just due to gravity acting on its mass. Using the coefficient of friction this produces a horizontal force of
-(19.62N*0.747)=-14.65N. From(2)&(7) : (8)
The sign is negative since it resists being dragged toward the pulley.

Due to gravity, Q is producing a force of 29.43N. From (1).

The total force on P and Q together is
29.43N - 14.65N = 14.78N. From (1)&(8) : (9)
----
F = ma so a = F/m = 14.78N/(3kg+2kg)
= (14.78kg-m/sec^2)/5kg = 2.956 m/sec^2. From(8) : (10)

Since P is 2kg and F = ma = 2kg * 2.956m/sec^2
= 5.912kg-m/sec^2 = 5.912N. From(10) : (11)

So with the friction on P and the force required to accelerate P, the tension in the string is
14.65N + 5.912N = 20.56N. From(8)&(11) : (12)
==============
We can double check by calculating the force due to gravity on Q that is left over based on it's acceleration.

If there were no tension in the string, in other words, nothing pulling up on Q it would accelerate down at 9.81m/sec^2. It is accelerating down at only 2.956m/sec^2, which is obviously the same as for P and Q together.

The resistance to being accelerated down is
3kg * 2.956m/sec^2 = 8.868N. From (10) : (13)

The tension in the string is the force of gravity on Q minus Q's resistance to being accelerated, which is
29.43N - 8.868N = 20.56N. From(1)&(13) : (14).

The double check is good, since both (12) and (14) produced a result of 20.56N.

gAr · 2011-02-28 21:34:51

Hi Kerrin,

byronjordan has done it correctly.
Here is another solution with free body diagrams of P and Q.

1)
Resolve the force along x-axis and y-axis. I assume the friction to be acting in the direction as shown.
At load P,

At load Q,

Substitute T and in

Since the answer is positive, our assumption of the force due to friction is correct.

2)
On removing the force of 30N, the load now moves to the right.
Let a be the acceleration of P and Q, m1 and m2 be the mass of Q and P respectively.
At load P,

At load Q,

Substituting for m1=3,m2=2, mu_k=0.746588287114035 and g=9.81,
T≈ 20.5608373159064N

Last edited by gAr (2011-02-28 21:35:51)

byronjordan · 2011-03-01 02:28:49

I was thinking a couple of things should be mentioned.
-----
The solution is based on the 30N angled up force on P, being the least force that prevents P from sliding. If the coefficient of friction were appreciably larger, a smaller force than 30N would still not allow P to slide. From the wording of the problem, 30N is obviously the least. 30N is in "exact balance" with the frictional force, allowing the coefficient to be calculated. This consideration is somewhat trivial.
-----
Also, as a general rule, static friction is greater than kinetic friction. In other words, the frictional force, when starting something sliding from a dead stop, is greater than the frictional force when it is sliding.

An example is automotive anti-lock brakes stopping a lot quicker than having all tires completely locked up and skidding. Both cars are moving until they stop, but with anti-lock brakes the tires aren't sliding on the pavement the entire time, therefore they are "static" relative to the pavement some of the time.

You might also experience this when trying to push something really heavy. That initial push can be a lot harder than the pushing needed to keep it going.

Again, from the wording of the problem, the same coefficient for the static and the kinetic condition is used. That's okay for solving a problem like this, but in the real world, that might not be the case.

Last edited by byronjordan (2011-03-01 02:31:46)

gAr · 2011-03-01 02:49:18

Hi byronjordan,

Yes, you're right.

And there's one book I'd like to mention for anybody interested in physics -
"Fundamentals of physics" by Jearl Walker, Halliday and Resnick

Math Is Fun Forum

#1 2011-02-28 07:59:36

Mechanics - Friction

#2 2011-02-28 15:41:27

Re: Mechanics - Friction

#3 2011-02-28 21:34:51

Re: Mechanics - Friction

#4 2011-03-01 02:28:49

Re: Mechanics - Friction

#5 2011-03-01 02:49:18

Re: Mechanics - Friction

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