In the figure below,
m1 = 10.9 kg
and
m2 = 3.7 kg.
The coefficient of static friction between m1 and the horizontal surface is 0.60, and the coefficient of kinetic friction is 0.30.
(a) If the system is released from rest, what will its acceleration be? (Enter the magnitude of the acceleration.)
m/s2
(b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system? (Enter the magnitude of the acceleration.)
m/s2
(a) Well, to find the acceleration of this system, we need to figure out the net force acting on it. Now, since m1 is resting on a horizontal surface, the force of static friction between m1 and the surface will be the maximum it can be, which is equal to the coefficient of static friction multiplied by the normal force. The normal force is just the weight of m1, which is m1 times the acceleration due to gravity. So, the force of static friction is just 0.60 times m1 times 9.8 m/s^2.
On the other hand, m2 is being pulled downward by the force of gravity. So the force we need to consider is just m2 times 9.8 m/s^2.
Now, since the system is released from rest, the net force is going to be equal to the force of static friction minus the force due to gravity.
So the net force is: (0.60*m1*9.8) - (m2*9.8).
Once we know the net force, we can figure out the acceleration using Newton's second law, F = ma. Rearranging the equation gives us a = F/m. Plugging in the values we know, we get:
a = ((0.60*m1*9.8) - (m2*9.8))/(m1 + m2).
Now you just need to plug in the values of m1 and m2 to find the acceleration. I don't have those values, so I can't give you the answer specifically, but I hope this helps!
(b) To find the acceleration of the system when m2 is moving downward, we need to consider the force of kinetic friction between m1 and the surface. The force of kinetic friction is equal to the coefficient of kinetic friction multiplied by the normal force. Again, the normal force is just the weight of m1, which is m1 times the acceleration due to gravity. So, the force of kinetic friction is just 0.30 times m1 times 9.8 m/s^2.
Now, since m2 is still being pulled downward by the force of gravity, the net force is going to be the force of kinetic friction minus the force due to gravity.
So the net force is: (0.30*m1*9.8) - (m2*9.8).
Once we know the net force, we can again use Newton's second law to find the acceleration using the equation a = F/m.
Plugging in the values we know, we get:
a = ((0.30*m1*9.8) - (m2*9.8))/(m1 + m2).
Again, I don't have the specific values for m1 and m2, so you'll have to plug those in yourself to find the magnitude of the acceleration. Good luck!
To answer both parts of the problem, we need to analyze the forces acting on the system.
First, let's consider when the system is released from rest.
(a) When the system is released from rest, the maximum force of static friction will act to prevent m2 from sliding. The force of static friction is given by:
F_static = coefficient of static friction * normal force
The normal force can be calculated as:
normal force = weight of m1 = m1 * g
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Then, we can use Newton's second law to determine the acceleration:
F_net = F_static - F_gravity
m1 * a = F_static - m1 * g
Since m1 is not moving, the force acting on it is only the force of static friction. Therefore, we can rewrite the equation as:
m1 * a = coefficient of static friction * normal force - m1 * g
Substituting the given values:
m1 = 10.9 kg
g = 9.8 m/s^2
coefficient of static friction = 0.60
m1 * a = 0.60 * (m1 * g) - m1 * g
a = 0.60 * g - g
a = (0.60 - 1) * g
a = -0.40 * g
Taking the magnitude of the acceleration:
a = 0.40 * g
a = 0.40 * 9.8 m/s^2
Therefore, the magnitude of the acceleration when the system is released from rest is:
0.40 * 9.8 = 3.92 m/s^2
(b) If the system is set in motion with m2 moving downward, the force of kinetic friction will act between m1 and the surface. The force of kinetic friction is given by:
F_kinetic = coefficient of kinetic friction * normal force
Using the same formula as above to determine the normal force, we can rewrite the equation for the acceleration:
m1 * a = F_kinetic - m1 * g
Substituting the given values:
m1 = 10.9 kg
g = 9.8 m/s^2
coefficient of kinetic friction = 0.30
m1 * a = 0.30 * (m1 * g) - m1 * g
a = 0.30 * g - g
a = (0.30 - 1) * g
a = -0.70 * g
Taking the magnitude of the acceleration:
a = 0.70 * g
a = 0.70 * 9.8 m/s^2
Therefore, the magnitude of the acceleration when the system is set in motion with m2 moving downward is:
0.70 * 9.8 = 6.86 m/s^2
To find the acceleration of the system in both scenarios, we need to consider the forces acting on each mass and use Newton's second law of motion.
(a) If the system is released from rest, the two masses will begin to accelerate because of the gravitational force acting on m2. Since m1 and m2 are connected, they will have the same acceleration. Let's assume the direction of acceleration is to the right.
To find the net force acting on each mass, we need to consider the forces involved. The forces acting on m1 are the normal force (N1) and the frictional force (f1) opposing its motion. The forces acting on m2 are the gravitational force (mg2) and the frictional force (f2) opposing its motion.
The equation for the net force on m1 is:
Net force on m1 = f1 = μs * N1
where μs is the coefficient of static friction between m1 and the surface and N1 is the normal force acting on m1.
The equation for the net force on m2 is:
Net force on m2 = mg2 - f2
where mg2 is the gravitational force acting on m2 and f2 is the kinetic frictional force acting on m2.
Since the magnitudes of the normal forces on m1 and m2 are equal to their weights (mg1 and mg2 respectively), we can rewrite the equations for net forces:
Net force on m1 = f1 = μs * mg1
Net force on m2 = mg2 - f2 = mg2 - μk * mg2
For both masses, the net force is equal to the mass times the acceleration:
Net force on m1 = m1 * a
Net force on m2 = m2 * a
Setting the equations equal to each other and solving for a:
μs * mg1 = (m1 + m2) * a
Plugging in the given values:
μs * m1 * g = (m1 + m2) * a
0.6 * 10.9 kg * 9.8 m/s^2 / 14.6 kg = a
Calculating this expression yields: a ≈ 4.07 m/s^2.
Therefore, the magnitude of the acceleration of the system when released from rest is approximately 4.07 m/s^2.
(b) If the system is set in motion with m2 moving downward, the normal force on m1 (N1) will be greater than the normal force on m2 (N2) due to the additional weight of m2. Now, let's assume the direction of acceleration is still to the right.
Again, we need to find the net forces acting on each mass. The equation for the net force on m1 remains the same:
Net force on m1 = f1 = μs * N1
The equation for the net force on m2 becomes:
Net force on m2 = mg2 - f2 = mg2 - μk * N2
Since the magnitudes of the normal forces on m1 and m2 are different, we can rewrite the equations for net forces:
Net force on m1 = f1 = μs * N1
Net force on m2 = mg2 - f2 = mg2 - μk * N2
Again, the net forces are equal to the respective masses times the acceleration:
Net force on m1 = m1 * a
Net force on m2 = m2 * a
Setting the equations equal to each other and solving for a:
μs * N1 = m2 * g - μk * N2
Since N1 = mg1 and N2 = mg2 - T (Tension force), we can rewrite the equation as follows:
μs * mg1 = m2 * g - μk * (mg2 - T)
We also know that the tension force, T, is equal to m1 * a (from Newton's second law).
We can substitute this into the equation and solve for a:
μs * mg1 = m2 * g - μk * (mg2 - m1 * a)
Plugging in the given values:
0.6 * 10.9 kg * 9.8 m/s^2 = 3.7 kg * 9.8 m/s^2 - 0.3 * (3.7 kg * 9.8 m/s^2 - 10.9 kg * a)
Calculating this expression yields: a ≈ 3.08 m/s^2.
Therefore, the magnitude of the acceleration of the system when set in motion with m2 moving downward is approximately 3.08 m/s^2.
I could guess I suppose, but better you describe situation.
if m2 is hanging
m2 g -T = m2 a or T = m2 g - m2 a
T - .6 m1 g = m1 a or T = .6 m1 g + m1 a
so
(m1+m2)a = (m2-.6 m1) g
( have a hunch it will not move. .6m1 >m2:)
do it all again with .3