In the figure below, m1 = 11.0 kg and m2 = 3.5 kg. The coefficient of static fri
ID: 2049144 • Letter: I
Question
In the figure below, m1 = 11.0 kg and m2 = 3.5 kg. The coefficient of static friction between m1 and the horizontal surface is 0.60 while the coefficient of kinetic friction is 0.30.
(a) If the system is released from rest, what will its acceleration be?
I got this answer of 0 m/s^2
(b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system?
For this answer I did the following:
f= kinetic friction x N
f= 0.30 (m)(g)
f=0.30(11)(9.8) = 32.3 N
T= 3.5 (g)
T=3.5(9.8) = 34.3N
F=T-f
F=34.3N-32.3N
a= F/ 11kg
a = 2N/ 11kg = .182
My answer is only 10% off, but I'm stuck on where I went wrong.
Explanation / Answer
(a) on M1=> Ffric = (.6)(11*9.81) = 64.76N on M2=> Fgrav = (9.81)(3.5) = 34.3N From this we can see that the frictional force acting on M1 (acting away from pulley) is greater than the force of gravity acting downwards on M2. Therefore, if this system was set up from rest, it would not accelerate, rather, it would remain static. (b) on M1=> Ffric = (0.3)(11*9.81) = 32.37N on M2 (from above) => Fgrav = 34.3 So from this, the force pulling on m2 is greater than the frictional force acting on m1, so this system will accelerate, with M1 accelerating toward the pulley, and M2 accelerating downward. But how much acceleration? Force = mass * acceleration. So the total force acting on the system = total mass * acceleration (34.3N -32.37N) = (3.5+11) * a Solving for a => a = 1.96/14.5 = 0.135 m/s^2