Three blocks rest on a frictionless, horizontal table (see figure below), with m1 = 8 kg and m3 = 20 kg. A horizontal force F = 101 N is applied to block 1, and the acceleration of all three blocks is found to be 3.5 m/s2.
Find m2.
What is the normal force between blocks 2 and 3?
F = (m1 + m2 + m3) a
You already know m1 and m3, a and F.
Solve for m2.
The force between blocks 2 and 3 is what acclerates m3. That force is m3*a
To find m2, we can use Newton's second law, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration (F = ma).
First, let's find the total mass of the system, which is the sum of the masses of all three blocks:
Total mass (m_total) = m1 + m2 + m3
Since the acceleration is the same for all three blocks, we can write the equation of motion for each block in terms of the total mass:
Block 1: F = m1 * a
Block 2: N2 - N3 = m2 * a
Block 3: N3 = m3 * a
Here, N2 and N3 represent the normal forces between blocks 2 and 1, and blocks 2 and 3, respectively.
From the second equation, we can see that N2 = F + N3. By substituting the values of F and N3, we get:
N2 = F + (m3 * a)
Now, let's substitute the values given:
F = 101 N
m1 = 8 kg
m3 = 20 kg
a = 3.5 m/s^2
We know that the acceleration of all three blocks is the same, so we can substitute the value of a into the above equation and solve for N2.
N2 = (F + (m3 * a))
N2 = (101 N + (20 kg * 3.5 m/s^2))
N2 = (101 N + 70 N)
N2 = 171 N
Therefore, the normal force between blocks 2 and 3 (N2) is equal to 171 N.
Now let's find the value of m2. We can use the equation of motion for block 2:
N2 - N3 = m2 * a
Substituting the values of N2 and N3, we get:
171 N - (m3 * a) = m2 * a
Substituting the given values, we can solve for m2:
171 N - (20 kg * 3.5 m/s^2) = m2 * 3.5 m/s^2
171 N - 70 N = m2 * 3.5 m/s^2
101 N = m2 * 3.5 m/s^2
Dividing both sides of the equation by 3.5 m/s^2, we get:
m2 = 101 N / 3.5 m/s^2
m2 ≈ 28.9 kg
Therefore, m2 is approximately 28.9 kg.
To solve this problem, we can use Newton's second law of motion and the concept of the net force acting on each block. Let's go step-by-step to find the values of m2 and the normal force between blocks 2 and 3.
Step 1: Identify the forces acting on each block:
- Block 1 experiences the applied force F and the normal force from block 2.
- Block 2 experiences the normal force from block 1 and block 3.
- Block 3 experiences the normal force from block 2 and its weight.
Step 2: Apply Newton's second law to each block:
- For block 1:
F_net1 = m1 * acceleration
F - normal force1 = m1 * acceleration ------ (Equation 1)
- For block 2:
F_net2 = m2 * acceleration
normal force1 - normal force2 = m2 * acceleration ------ (Equation 2)
- For block 3:
F_net3 = m3 * acceleration
normal force2 - m3 * g = m3 * acceleration ------ (Equation 3)
Step 3: Solve the equations simultaneously to find m2 and the normal force2.
From Equation 1:
F - normal force1 = m1 * acceleration
101 N - normal force1 = 8 kg * 3.5 m/s²
101 N - normal force1 = 28 N ------ (Equation 4)
From Equation 2:
normal force1 - normal force2 = m2 * acceleration
normal force1 - normal force2 = m2 * 3.5 m/s² ------ (Equation 5)
From Equation 3:
normal force2 - m3 * g = m3 * acceleration
normal force2 - 20 kg * 9.8 m/s² = 20 kg * 3.5 m/s²
normal force2 - 196 N = 70 N
normal force2 = 266 N ------ (Equation 6)
Step 4: Substitute the values from Equation 6 into Equation 5:
normal force1 - normal force2 = m2 * 3.5 m/s²
28 N - 266 N = m2 * 3.5 m/s²
m2 = -238 N / (3.5 m/s²)
m2 ≈ -68 kg
Step 5: Analyze the result:
The negative value for m2 suggests an error in the calculations or the question's setup. It is not physically possible to have a negative mass. Please double-check the problem setup and values provided to ensure accuracy.
Regarding the normal force between blocks 2 and 3, we found that it is 266 N in Equation 6.