In the previous quiestion the minimum force F required so that block A will slip on block B is:

Options

A. 30 N
B. 20 N
C. 10 N
D. not possible

Explanation

To solve the problem of determining the minimum force \( F \) required for block A to slip on block B, we can follow these steps:

### Step 1: Understand the System
We have two blocks: Block A (10 kg) resting on Block B (20 kg). We need to find the minimum force \( F \) that needs to be applied to Block A so that it starts to slip over Block B.

### Step 2: Calculate the Normal Force on Block A
The normal force \( N \) acting on Block A is equal to its weight since it is resting on Block B.

\[ N_A = m_A \cdot g = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \]

### Step 3: Calculate the Maximum Static Friction Force between A and B
The maximum static friction force \( f_{\text{max}} \) that can act between Block A and Block B is given by:

\[ f_{\text{max}} = \mu_s \cdot N_A \]

Where \( \mu_s \) is the coefficient of static friction between the two blocks. Given \( \mu_s = 0.2 \):

\[ f_{\text{max}} = 0.2 \cdot 100 \, \text{N} = 20 \, \text{N} \]

### Step 4: Determine the Minimum Force Required
For Block A to start slipping over Block B, the applied force \( F \) must exceed the maximum static friction force. Thus, the minimum force \( F \) required is:

\[ F_{\text{min}} = f_{\text{max}} = 20 \, \text{N} \]

### Conclusion
The minimum force \( F \) required for Block A to slip on Block B is \( 20 \, \text{N} \).

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