A block of mass m1 is on top of a block of mass m2. Block 2 is connected by an ideal rope passing through a pulley to a block of unknown mass m3 as shown. The pulley is massless and frictionless. There is friction between block 1 and 2 and between the horizontal surface and block 2. Assume that the coefficient of kinetic friction between block 2 and the surface, μ, is equal to the coefficient of static friction between blocks 1 and 2.
a) What is the value of m3 for which m1 and m2 are moving together at a constant speed? Express your answer in terms of some or all of the variables m1, m2, μ, and g. To write μ, use mu.
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b) The mass of block 3 has been changed such that block 1 and block 2 are moving together with a given acceleration of magnitude a. What is the magnitude and the direction of the force of friction exerted by block 2 on block 1? Express your answer in terms of some or all of the variables a, m1, m2, μ, and g. To indicate the direction, use a minus sign if the force is pointing to the left. To write μ, use mu.
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c) What is the minimum value of m3 for which block 1 will start to move relative to block 2? Express your answer in terms of some or all of the variables m1, m2, μ, and g. To write μ, use mu.
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For a)
mu*(m1+m2)
for b) m1 * a1
a) To find the value of m3 for which m1 and m2 are moving together at a constant speed, we need to consider the forces acting on the system.
Since m1 and m2 are moving together, their acceleration is the same. Let's denote this common acceleration as a.
The force pulling block 1 is the tension in the rope, T. The force pulling block 2 is also T since the rope is massless and frictionless. The force of friction between block 1 and block 2 is μ times the normal force, which is equal to m2 times g (where g is the acceleration due to gravity).
Applying Newton's second law to each block:
m1 * a = T - μ * (m2 * g) [1]
m2 * a = T - μ * (m1 * g) [2]
Since we want the blocks to move together at a constant speed, the acceleration should be zero. Setting a = 0 in equations [1] and [2], we get:
T = μ * (m1 * g) [3]
T = μ * (m2 * g) [4]
From equations [3] and [4], we can see that T is the same in both cases. Therefore, we can equate the equations:
μ * (m1 * g) = μ * (m2 * g)
Simplifying and rearranging, we find:
m1 = m2
Now, substituting m1 = m2 into equations [3] or [4], we get:
T = μ * (m1 * g)
Since T = m3 * a, we can get the value of m3:
m3 = (μ * (m1 * g)) / a
Therefore, the value of m3 for which m1 and m2 are moving together at a constant speed is given by:
m3 = (μ * (m1 * g)) / a
b) To find the magnitude and direction of the force of friction exerted by block 2 on block 1 when they are moving together with acceleration a, we can consider the forces acting on block 2.
The force pulling block 2 is T, which is equal to m3 * a. The force of friction between block 2 and the surface is μ times the normal force, which is equal to m2 * g.
Applying Newton's second law to block 2:
m2 * a = μ * (m2 * g) - T
Substituting T = m3 * a, we get:
m2 * a = μ * (m2 * g) - m3 * a
Rearranging the equation, we find:
m3 * a = μ * (m2 * g) - m2 * a
Simplifying further, we get:
m3 * a + m2 * a = μ * (m2 * g)
Combining like terms, we have:
(m3 + m2) * a = μ * (m2 * g)
Finally, we can find the magnitude and direction of the force of friction by multiplying both sides of the equation by -1 (since the force of friction opposes the motion):
Magnitude of friction force = μ * (m2 * g) - (m3 + m2) * a
Direction of friction force: To the left (opposite direction of motion)
c) The minimum value of m3 for which block 1 will start to move relative to block 2 can be found by considering the forces acting on the system when it is on the verge of motion.
In this case, the force of static friction between block 1 and block 2 will be at its maximum value, which is μ times the normal force, equal to m2 * g.
Applying Newton's second law to block 1:
m1 * a = μ * (m2 * g)
Simplifying, we find:
m3 = μ * (m2 * g) / a
Therefore, the minimum value of m3 for which block 1 will start to move relative to block 2 is given by:
m3 = μ * (m2 * g) / a
To solve these problems, we can use Newton's second law and the concepts of force and friction. Let's break down each part one by one.
a) To find the value of m3 for which m1 and m2 are moving together at a constant speed, there are two key forces to consider: the force of gravity acting on each block, and the force of friction between block 1 and block 2. Since the blocks are moving together at a constant speed, the acceleration is zero.
First, let's consider the forces acting on m1:
- The force of gravity acting on m1 is given by F1 = m1 * g, where g is the acceleration due to gravity.
- The force of friction acting on m1 is equal to the coefficient of static friction (μ) multiplied by the normal force between m1 and m2, which is equal to F_n = m1 * g. So, F_f1 = μ * F_n = μ * (m1 * g).
Now, let's consider the forces acting on m2:
- The force of gravity acting on m2 is given by F2 = m2 * g.
- The force of friction acting on m2 is the same as the force acting on m1, so F_f2 = μ * (m1 * g).
Since the two blocks are moving together, the tension in the rope connecting them is the same. Let's denote the tension as T.
Now, we can write the equation of motion for the system:
T - F_f2 = m2 * a -- (1) (where 'a' is the acceleration of the system)
F_f1 - T = m1 * a -- (2)
Since the blocks are moving at a constant speed, the acceleration is zero. Therefore, we can set a = 0 in equations (1) and (2):
T - F_f2 = 0 -- (3)
F_f1 - T = 0 -- (4)
Substituting the values of F_f1 and F_f2 from above, we have:
μ * (m1 * g) - T = 0 -- (5)
T - μ * (m1 * g) = 0 -- (6)
We can solve equations (5) and (6) simultaneously to find the value of T and thus m3. Adding equations (5) and (6), we get:
T = μ * (m1 * g) / 2
Since the tension in the rope is the same as the force of friction between the two blocks, we can equate T to μ * (m1 * g) / 2:
μ * (m1 * g) / 2 = μ * (m1 * g)
Simplifying, we find:
m3 = 2 * m1
Therefore, the value of m3 for which m1 and m2 are moving together at a constant speed is m3 = 2 * m1.
b) The magnitude and direction of the force of friction exerted by block 2 on block 1 can be found by using Newton's second law. Since block 1 and block 2 are moving together with a given acceleration (a), we can relate the force of friction (F_f1) with the known forces acting on block 1:
Using the equation F_f1 = m1 * a, we substitute the value of a from the given data.
c) To find the minimum value of m3 for which block 1 will start to move relative to block 2, we need to determine the maximum force of friction (F_f1) that block 2 can exert on block 1 without causing block 1 to move. This occurs when static friction is at its maximum value, which is equal to μ * F_n, where F_n is the normal force between block 1 and block 2.
The normal force (F_n) is equal to the force of gravity acting on block 1, which is m1 * g. So, we have:
F_f1 (max) = μ * (m1 * g)
Since the blocks are on the verge of moving, the friction force must be equal to the maximum friction force:
F_f1 (max) = F_f2 = μ * (m2 * g)
Setting these two equations equal to each other and solving for m3, we get:
μ * (m1 * g) = μ * (m2 * g)
m1 = m2 * (m3 + m2) / m3
Simplifying, we find:
m3 = (m1 * m2) / (m2 - m1)
Therefore, the minimum value of m3 for which block 1 will start to move relative to block 2 is m3 = (m1 * m2) / (m2 - m1).