Question

Two blocks A and B of mass 120 N $m_A$ = 10 kg and $m_B$ = 20 kg are place on rough horizontal surface. The blocks are connected with a string. If the coefficient of friction between block A and ground is $\mu_A$ = 0.9 and between block B and ground is $\mu_B$ = 0.3, find the tension in the string in situation as shown in figure. Forces 120 N and 100 N start acting when the system is at rest?

Answer 1

40 N

Answer 2

To find the tension in the string, we first need to determine if the system of blocks will move or remain at rest.

1. Calculate Maximum Static Friction for each block: Let $g = 10 \, \text{m/s}^2$.

• For Block A: Mass $m_A = 10 \, \text{kg}$ Normal force $N_A = m_A g = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N}$ Coefficient of friction $\mu_A = 0.9$ Maximum static friction $f_{s,max,A} = \mu_A N_A = 0.9 \times 100 \, \text{N} = 90 \, \text{N}$

• For Block B: Mass $m_B = 20 \, \text{kg}$ Normal force $N_B = m_B g = 20 \, \text{kg} \times 10 \, \text{m/s}^2 = 200 \, \text{N}$ Coefficient of friction $\mu_B = 0.3$ Maximum static friction $f_{s,max,B} = \mu_B N_B = 0.3 \times 200 \, \text{N} = 60 \, \text{N}$

2. Determine the tendency of motion for the entire system:

• Applied force on A, $F_A = 120 \, \text{N}$ (to the left).
• Applied force on B, $F_B = 100 \, \text{N}$ (to the right).

Consider the system of blocks (A+B) as a single unit. The net external horizontal force acting on the system is: $F_{net} = F_A - F_B = 120 \, \text{N} - 100 \, \text{N} = 20 \, \text{N}$ (to the left).

The total maximum static friction that can oppose this net force is the sum of maximum static frictions on each block: $f_{s,max,total} = f_{s,max,A} + f_{s,max,B} = 90 \, \text{N} + 60 \, \text{N} = 150 \, \text{N}$.

Since the net applied external force ($20 \, \text{N}$) is less than the total maximum static friction ($150 \, \text{N}$), the system will remain at rest. Therefore, the acceleration of the system $a = 0$.

3. Apply Equilibrium Conditions to each block: Since the system is at rest, the net force on each block is zero. Let $T$ be the tension in the string.

• Free Body Diagram for Block A: Forces acting horizontally on Block A:

  • $F_A = 120 \, \text{N}$ (to the left)
  • Tension $T$ (to the right)
  • Static friction $f_A$ (to the right, opposing the tendency of motion to the left)

  Applying Newton's First Law ($\Sigma F_x = 0$): $F_A - T - f_A = 0$ $120 - T - f_A = 0 \quad (1)$

• Free Body Diagram for Block B: Forces acting horizontally on Block B:

  • $F_B = 100 \, \text{N}$ (to the right)
  • Tension $T$ (to the left)
  • Static friction $f_B$ (to the left, opposing the tendency of motion to the right)

  Applying Newton's First Law ($\Sigma F_x = 0$): $F_B - T - f_B = 0$ $100 - T - f_B = 0 \quad (2)$

4. Solve for Tension (T): From equation (1), $T = 120 - f_A$. From equation (2), $T = 100 - f_B$.

Equating the two expressions for T: $120 - f_A = 100 - f_B$ $f_A - f_B = 20 \quad (3)$

We know that $0 \le f_A \le f_{s,max,A} = 90 \, \text{N}$ and $0 \le f_B \le f_{s,max,B} = 60 \, \text{N}$.

Since Block B has a greater tendency to move (its applied force $F_B=100$ N is greater than its $f_{s,max,B}=60$ N, meaning it would move if isolated), it will utilize its maximum static friction first to resist motion. Let's assume block B is on the verge of slipping, so $f_B = f_{s,max,B} = 60 \, \text{N}$.

Substitute $f_B = 60 \, \text{N}$ into equation (3): $f_A - 60 = 20$ $f_A = 80 \, \text{N}$

Now, check if this value of $f_A$ is within its maximum limit: $80 \, \text{N} \le 90 \, \text{N}$. Yes, it is valid. Since both static friction values ($f_A = 80 \, \text{N}$ and $f_B = 60 \, \text{N}$) are within their maximum possible values, the assumption that the system remains at rest is consistent.

Finally, substitute the value of $f_B$ (or $f_A$) back into the tension equations: Using equation (2): $100 - T - 60 = 0$ $40 - T = 0$ $T = 40 \, \text{N}$

(As a check, using equation (1): $120 - T - 80 = 0 \implies 40 - T = 0 \implies T = 40 \, \text{N}$. Both yield the same result.)

The tension in the string is 40 N.