Questions: Three blocks are connected on the table as shown below. The table is rough and has a coefficient of kinetic friction of 0.440. The objects have masses of 4.00 kg, 1.00 kg, and 2.00 kg, as shown, and the pulleys are frictionless. (a) Draw free-body diagrams for each of the objects. (Do this on paper. Your instructor may ask you to turn in this work.) (b) Determine the acceleration of each object and their directions. (c) Determine the tensions in the two cords.

Solution Steps

For the 4 kg mass (hanging on the left), the forces acting are the tension in the left cord (T₁) upwards and its weight (m₁g) downwards.

For the 1 kg mass (on the table), the forces acting are the tension in the left cord (T₁) to the left, the tension in the right cord (T₂) to the right, the normal force (N) upwards, and its weight (m₂g) downwards, and the frictional force (μN) to the right.

For the 2 kg mass (hanging on the right), the forces acting are the tension in the right cord (T₂) upwards and its weight (m₃g) downwards.

Applying Newton's second law to each mass:

• 4 kg mass: T₁ - m₁g = -m₁a (Downward acceleration)
• 1 kg mass: T₂ - T₁ - μm₂g = m₂a (Rightward acceleration)
• 2 kg mass: T₂ - m₃g = -m₃a (Downward acceleration)

Here, 'a' is the magnitude of the acceleration and is the same for all masses since they are connected. μ is the coefficient of kinetic friction.

We have three equations and three unknowns (T₁, T₂, and a).

First, find the normal force on the 1 kg block: N = m₂g.

Next, substitute N into the equation for the 1 kg block: T₂ - T₁ - μm₂g = m₂a

Now we have three equations:

1. T₁ - 4g = -4a
2. T₂ - T₁ - μg = a
3. T₂ - 2g = -2a

We can solve this system of equations. Adding equations (1) and (2) eliminates T₁, giving: T₂ - (4g + μg) = -3a

From equation (3), T₂ can be expressed as: T₂ = 2g - 2a

Substitute this expression for T₂ into the equation T₂ - (4g + μg) = -3a :

2g - 2a - (4g + μg) = -3a

Simplify and solve for 'a': a = (2 + μ)g

a = (2 + 0.44) * 9.8 = 23.872 m/s²

Now, substitute 'a' back into the equations to find the tensions.

T₁ = 4g - 4a = 4(9.8) - 4(23.872) ≈ -56.688 N. Since tension cannot be negative, the correct value of acceleration must have been different from that which we got earlier. Let us correct our approach.

If we add equations (1), (2), and (3), we get the following: -m₁g - μm₂g - m₃g = -m₁a + m₂a -m₃a. Solving for a, we get a = g(m₁ + μm₂ + m₃)/(m₁ + m₂ + m₃) = 9.8(4 + 0.44 + 2)/(4+1+2) = 8.25 m/s².

Now T₂ = 2g - 2a = 2(9.8) - 2(8.25) = 3.1 N.

Also, T₁ = 4g - 4a = 4(9.8) - 4(8.25) = 6.2 N

Final Answer: