A string passes over a frictionless, massless pulley attached to the ceiling. A mass m1=15.0 kg hangs from one end of this string, and a second massless, frictionless pulley hangs from the other end of the string. A second string passes over the second pulley and a mass m2=4.0 kg hangs from the end, whereas the other end is attached firmly to the ground. Assume positive accelerations are upward.
1). What is the acceleration of m1?
2). What is the acceleration of m2?
3). What is the acceleration of the second pulley?
WHAT I TRIED:
To find m1 I tried a = (m1-m2)/(m2+m1) but that did not work. ( The value is negative but I entered it as positive that might be a problem)
To find m2 I tried m2a2=T-m2g, finding the tension of then implicating it in the formula to solve for the a2 acceleration but that did not work.
The last part I wrote 0 just because the second pulley does not have mass, so I need help.
Letting a be the acceleration of m1, and b be the acceleration of m2.
Note that the lower pulley will have an accelerlation of b/2, but this is the acceleration of m2. So
a=b/2
Consider the force on the ceiling holding the upper pulley:
F=m1(g-a)+m2(g+a)
and the tension on the lower rope..
T=m2(g+b)
But the upper force -lower force=system mass* system acceleration
or F-T=m1*a-m2*b
or
m1(g-a)+m2(g+a)-m2(g+b)=m1*a-m2*b
Think on that, it is after all, three am here, and I could be in error.
If correct, you can solve for a, and b.
To determine the accelerations of m1, m2, and the second pulley, we can analyze the forces acting on each object and apply Newton's second law of motion.
1) To find the acceleration of m1:
The only force acting on m1 is its weight (mg). Considering upward as positive, the tension in the string (T) also acts upward. According to Newton's second law, the net force on m1 is equal to the product of its mass and acceleration (F_net = m1 * a1).
Since the pulley is frictionless and massless, the tension in the string is the same on both sides. Therefore, T = m2 * g, where g is the acceleration due to gravity.
Summing the forces on m1:
F_net = T - mg
m1 * a1 = T - m1 * g
m1 * a1 = m2 * g - m1 * g
a1 = (m2 - m1) * g / m1
a1 = (4.0 kg - 15.0 kg) * 9.8 m/s^2 / 15.0 kg
a1 ≈ -5.92 m/s^2 (upward acceleration)
2) To find the acceleration of m2:
The force acting on m2 is its weight (mg). The tension in the string (T) acts downward. Considering upward as positive, the net force on m2 is given by F_net = m2 * a2.
Considering the second pulley, its acceleration will be half of the acceleration of m2, as it moves twice the distance compared to m2.
Summing the forces on m2:
F_net = mg - T
m2 * a2 = m2 * g - T
m2 * a2 = m2 * g - m2 * g
a2 = 0 m/s^2 (m2 does not accelerate)
3) Since the second pulley does not have mass, it will have the same acceleration as m2: a2 = 0 m/s^2. In this case, the second pulley remains at rest.
In summary:
1) The acceleration of m1 is approximately -5.92 m/s^2 (upward).
2) The acceleration of m2 is 0 m/s^2 (no acceleration).
3) The acceleration of the second pulley is also 0 m/s^2 (no acceleration).
To solve this problem, we can apply Newton's second law of motion to each object involved.
1. Acceleration of m1:
Let's consider the forces acting on m1. There is tension (T) acting upward and the force due to gravity (m1 * g) acting downward. Since the positive direction of acceleration is upward, we can write the equation:
m1 * a = T - m1 * g
Now, we need to find the tension (T) in terms of the given variables. Since the string is massless and frictionless, the tension is the same throughout the string. Therefore, the tension in the string attached to m1 also applies to the other end attached to the second pulley.
We can represent the tension in terms of the mass m2 and the acceleration of m2 (a2). From the forces acting on m2, we have:
m2 * a2 = m2 * g - T
Since the pulley is massless and frictionless, the tension on both sides of the pulley is the same. This means T = T. Therefore, we can equate these two equations to solve for T:
T = m2 * g - m2 * a2
Now, substitute this expression for T back into the equation for m1:
m1 * a = m2 * g - m2 * a2 - m1 * g
Simplifying further, we get:
a = (m2 * g - m2 * a2 - m1 * g) / m1
2. Acceleration of m2:
Using the equation for tension T, we have:
T = m2 * g - m2 * a2
Since the tension in the string attached to m1 is also T, we can substitute this expression into the equation for m1:
m1 * a = T - m1 * g
m1 * a = (m2 * g - m2 * a2) - m1 * g
Simplifying further, we get:
a = (m2 * g - m2 * a2 - m1 * g) / m1
As you can see, the accelerations of m1 and m2 are given by the same expression.
3. Acceleration of the second pulley:
Since the second pulley has no mass, it does not contribute to the overall acceleration. Therefore, the acceleration of the second pulley is 0.