In the figure, we have block B (with weight Wb) hanging from the pulley connected by a wire to block A (with weight Wa). If we were to let it start moving with constant speed, block B would go down and block A will move to the right. Friction is present between block A and the surface. We are tasked to find the coefficient of friction between block A and the surface.
(NOTE: We may isolate the system by parts. Here, I am taking the system as a whole)
First, we look at the forces involved.
- Weight of block B
- Tension (above block B)
- Tension (beside block A)
- Friction (on block A and surface)
Equations:
- F = ma = 0 (Since we have constant velocity, our acceleration is zero.)
We have the weight of block B (assumed positive) which goes downward and tension going upward, tension going the opposite direction of the weight of block B, and friction going against the weight of block B. So our equation becomes
- (mass of block B * gravity) – Tension(B) – Tension(A) – friction = 0
But wait! We lack tension. Recall that If we isolate block B and its tension, we can find for tension.
- (mass of block B * gravity) – Tension(B) = 0
- Or we can write this as Tension(B) = Weight(B)
Mass of block B times the gravity (or weight of block B) equals the tension of the string above block B. Also, Tension B equals Tension A (assuming the pulley is frictionless).
So our equation becomes
- (weight of block B) – Tension (B) – Tension (B) – friction = 0
- OR: weight of block B – weight of block B – weight of block B – friction = 0
- But in this example, let us use (weight of block B) – (2 * Tension) – friction = 0
Friction equals u(mew) times the normal force of block A. Thus,
- (weight of block B) – (2 * Tension) – u*normal force of block A = 0
The coefficient of friction (u) can be taken by the equation
- u = [(weight of block B) - (2*Tension)] / normal force of block A
By now, you should be able to get the coefficient of friction. Simply substitute the weights and tensions in the equations and you will arrive to your final answer.
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If you are to accelerate the system (say block B going downward), you will need Newton’s Second Law of Motion. The acceleration for block B is similar to the acceleration of block A because we assume a frictionless pulley (but not a frictionless surface).
- F = ma
In terms of the y-component (which only block B has),
- (mass of block B * g ) – Tension = mass of block B * acceleration
In the x-component (block A),
- Tension – friction = mass of block A * acceleration
The tension and friction are subtracted. If they are added, then the motion will be going to the left instead of the right. Also, if we add them, the wire (tension A) will break.
Using the equation Fnet = ma = Fa + Fb
- Fnet = (mass of block A *acceleration) + (mass of block B *acceleration)
- Fb = (mass of block B * gravity) – Tension = mass of block B * acceleration
- Fa = Tension – friction = mass of block A * acceleration
As mentioned earlier, their acceleration is the same. By combining equations, we get
(mass of block A * acceleration) + (mass of block B * acceleration) = [Tension - friction] +
[(mass of block B * gravity) - Tension]
Or we can factor out acceleration
a (mass of block A + mass of block B) = T – f + weight of block B – T (where tension =0)
In getting acceleration,
a = [weight of block B - friction] / [mass of block A +mass of block B]
By now, you should arrive at the proper value of acceleration. If you get a negative answer, it just means that it is decelerating.
