Question

The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is __________ N.
(take g=10 ms\(^{-2}\))
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Hint:

In problems with stacked blocks, always identify the force that links their motion (here, it's friction). Find the maximum acceleration that this linking force can provide to the block it's acting on. This will be the maximum acceleration for the entire system to move as one unit.

Answer 1

Correct Answer: 15

Step 1: Understanding the Question:
We have two blocks, one on top of the other, on a smooth horizontal table. A horizontal force F is applied to the lower block. We need to find the maximum force F that can be applied such that both blocks move together without the top block slipping.
Step 2: Key Formula or Approach:
1. For the blocks to move together, they must have the same acceleration, `a`.
2. The force causing the top block (m\(_1\) = 1 kg) to accelerate is the force of static friction, f\(_s\), exerted by the bottom block.
3. The maximum possible static friction is f\(_{s,max}\) = \(\mu_s N_1\), where N\(_1\) is the normal force on the top block. This maximum friction corresponds to the maximum possible acceleration, a\(_{max}\), for the blocks to move together.
4. Apply Newton's second law to the combined system (m\(_1\) + m\(_2\)) to find the maximum applied force F\(_{max}\).
Step 3: Detailed Explanation:
Let m\(_1\) = 1 kg and m\(_2\) = 2 kg. The coefficient of static friction \(\mu_s\) = 0.5.
First, consider the top block (m\(_1\)). The only horizontal force acting on it is the static friction from the bottom block. The free-body diagram for the top block gives:
\[ f_s = m_1 a \] The maximum acceleration occurs when the static friction is at its maximum value:
\[ f_{s,max} = \mu_s N_1 \] The normal force N\(_1\) on the top block is equal to its weight, N\(_1\) = m\(_1\)g.
\[ f_{s,max} = \mu_s m_1 g = 0.5 \times 1 \text{ kg} \times 10 \text{ m/s}^2 = 5 \text{ N} \] Now, we can find the maximum acceleration a\(_{max}\) that the top block can sustain:
\[ a_{max} = \frac{f_{s,max}}{m_1} = \frac{5 \text{ N}}{1 \text{ kg}} = 5 \text{ m/s}^2 \] For the blocks to move together, the entire system must accelerate at this maximum value, a\(_{max}\).
Now, consider the two blocks as a single system of total mass M = m\(_1\) + m\(_2\) = 1 kg + 2 kg = 3 kg. The external horizontal force is F. Applying Newton's second law to the system:
\[ F_{max} = M \cdot a_{max} = (m_1 + m_2) a_{max} \] \[ F_{max} = 3 \text{ kg} \times 5 \text{ m/s}^2 = 15 \text{ N} \] Step 4: Final Answer:
The maximum horizontal force that can be applied to move the blocks together is 15 N.