Three blocks are in contact with each other on a frictionless, horizontal surface, as shown below. A horizontal force is applied to m1. Take m1 = 2.00 kg, m2 = 3.00 kg, m3 = 5.00 kg, and F = 22.0 N
Find the acceleration of the blocks
_______m/s2 (to the right)
(c) Find the resultant force on m1.
_______N (to the right)
Find the resultant force on m2.
_______ N (to the right)
Find the resultant force on m3.
_______ N (to the right)
(d) Find the magnitude of the contact force between m1 and m2.
_______N
Find the magnitude of the contact force between m2 and m3.
_______ N
(e) You are working on a construction project. A coworker is nailing up plasterboard on one side of a light partition, and you are on the opposite side, providing "backing" by leaning against the wall with your back pushing on it. Every blow makes your back sting. The supervisor helps you to put a heavy block of wood between the wall and your back. Using the situation analyzed in parts (b), (c), and (d) as a model, explain how this works to make your job more comfortable.
_______________________________________
___________________________________
To find the acceleration of the blocks, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = ma).
Here's how to calculate the acceleration:
1. Calculate the total mass of the blocks by summing the individual masses: m_total = m1 + m2 + m3 = 2.00 kg + 3.00 kg + 5.00 kg = 10.00 kg
2. Calculate the net force on the system using the applied force F: F_net = F
3. Use Newton's second law to find the acceleration: F_net = m_total * a
Rearranging the equation, we get: a = F_net / m_total
Calculate the acceleration: a = 22.0 N / 10.00 kg = 2.20 m/s^2 (to the right)
Now let's move on to finding the resultant forces on the blocks:
(c) The resultant force on m1:
Since m1 only experiences the applied force F, the resultant force on m1 is equal to the applied force: Resultant force on m1 = F = 22.0 N (to the right)
The resultant force on m2:
Since m2 is in contact with both m1 and m3, it experiences the contact forces. The contact force between m1 and m2 and the contact force between m2 and m3 are equal and opposite due to Newton's third law.
(d) The resultant force on m3:
Since m3 only experiences the contact force from m2, the resultant force on m3 is equal to the contact force from m2.
To find the magnitude of the contact forces between the blocks:
(d) The magnitude of the contact force between m1 and m2:
Since m1 and m2 are in contact, the contact force between them is equal and opposite due to Newton's third law. Therefore, the magnitude of the contact force between m1 and m2 is equal to the magnitude of the contact force between m2 and m1.
(e) The magnitude of the contact force between m2 and m3:
Since m2 and m3 are in contact, the contact force between them is equal and opposite due to Newton's third law.
Now let's address the construction project scenario:
In this scenario, the heavy block of wood acts as an additional mass (similar to m1, m2, and m3) between the wall and your back. This additional mass increases the total mass of the system and, therefore, reduces the overall acceleration. As a result, when you lean against the wall with the heavy block of wood providing backing, the force exerted on your back (due to hammer blows on the plasterboard) is distributed over a larger mass, reducing the overall acceleration and making your job more comfortable by reducing the stinging sensation.
To find the acceleration of the blocks, we can apply Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. Since the blocks are in contact and experiencing the same acceleration, the net force applied to each block can be determined by considering each block separately.
(a) Acceleration of the blocks:
We'll start by considering m1 and m2 as a single system. The net force acting on this system is the force applied (F) minus the force of friction between m1 and m2. The frictional force can be found using the formula: frictional force = coefficient of friction * normal force. However, since the surface is frictionless, the force of friction is zero. Therefore, the net force acting on this system is simply the applied force (F).
m1 and m2 have a combined mass of 2.00 kg + 3.00 kg = 5.00 kg. Applying Newton's second law, we have:
F = (m1 + m2) * a
22.0 N = (2.00 kg + 3.00 kg) * a
22.0 N = 5.00 kg * a
a = 4.4 m/s² (to the right)
Now, considering m2 and m3 as a single system, the force applied (F) is the resultant force acting on m2, which is equal to the force of friction between m2 and m3. Again, since the surface is frictionless, the force of friction is zero. Therefore, the net force acting on this system is also simply the applied force (F).
m2 and m3 have a combined mass of 3.00 kg + 5.00 kg = 8.00 kg. Applying Newton's second law, we have:
F = (m2 + m3) * a
22.0 N = (3.00 kg + 5.00 kg) * a
22.0 N = 8.00 kg * a
a = 2.75 m/s² (to the right)
(b) Resultant force on m1:
Since m1 is experiencing the same acceleration as the other blocks, the resultant force acting on m1 is also equal to the applied force (F).
Resultant force on m1 = 22.0 N (to the right)
(c) Resultant force on m2:
As mentioned earlier, the resultant force acting on m2 is equal to the applied force (F).
Resultant force on m2 = 22.0 N (to the right)
(d) Resultant force on m3:
Similarly, the resultant force acting on m3 is also equal to the applied force (F).
Resultant force on m3 = 22.0 N (to the right)
(e) In this construction project scenario, when you lean against the wall, your back exerts a force on the wall equal to your weight. This force is transmitted through the wall and is experienced by your coworker on the other side as a reaction force. When a heavy block of wood is placed between the wall and your back, it increases the normal force between you and the wall. This increased normal force results in an increased frictional force between you and the wall. The increased friction helps to counterbalance the force applied by your coworker while nailing the plasterboard, reducing the sting you feel on your back. By increasing the normal force and friction, the block of wood provides additional support and stability, making your job more comfortable.