Question

A 0.5kg wooden block is placed on top of a 1.0kgwooden block. The coefficient static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.20 wht is the maximum horizontal force that can be applied to the lower block

Answer 1

A block of 0.5 kg is placed on top of another wooden block which weighs 1.0 kg. The coefficient of static friction between the two blocks is 0.35, whereas the coefficient of kinetic friction between the lower block and the level table is 0.20.

To calculate the maximum horizontal force that can be applied to the lower block, we need to determine the limiting frictional force between the two blocks.

Since the upper block is not moving, the force of static friction is acting on it. We can calculate this force as follows:

`F_static = friction coefficient * normal force`

where, normal force = weight of upper block = 0.5 kg * 9.81 m/s^2 = 4.905 N

`F_static = 0.35 * 4.905 = 1.718 N`

Therefore, the static frictional force acting on the upper block is 1.718 N.

Now, we need to find the maximum force that can be applied to the lower block before it starts moving. This force is equal to the force of static friction acting on the lower block.

Since the upper block is not moving, the force of static friction acting on the lower block is equal to the force of static friction acting on the upper block.

`F_static(lower block) = F_static(upper block) = 1.718 N`

This means that the maximum horizontal force that can be applied to the lower block is 1.718 N.

However, if the applied force exceeds this value, the lower block will start moving and the force of kinetic friction will be acting on it, which is equal to:

`F_kinetic = friction coefficient * normal force`

`F_kinetic = 0.20 * 4.905 = 0.981 N`

Hence, if the applied force exceeds 1.718 N, the lower block will start moving and the force of kinetic friction will act on it, which is 0.981 N.

Therefore, the maximum horizontal force that can be applied to the lower block is 1.718 N.

Answer 2

The maximum horizontal force that can be applied to the lower block without the upper block slipping is approximately 1.715 N.

Here's how to find the maximum horizontal force that can be applied to the lower block without the upper block slipping:

**1. Identify the forces involved:**

* **Weight of the upper block (W1):** 0.5 kg * 9.81 m/s² ≈ 4.9 N (directed downward)

* **Normal force (N):** exerted by the lower block on the upper block (directed upward)

* **Static friction force (fs):** opposes the applied force and prevents the upper block from slipping (directed opposite to the applied force)

* **Weight of the lower block (W2):** 1.0 kg * 9.81 m/s² ≈ 9.81 N (directed downward)

* **Kinetic friction force:** opposes the applied force once the upper block starts slipping (directed opposite to the applied force)

* **Applied force (F):** horizontal force applied to the lower block (directed to the right)

**2. Analyze the conditions for static equilibrium:**

Before the upper block starts slipping, the net force acting on it must be zero. Therefore:

* N - W1 = 0

**3. Relate normal force and static friction force:**

The maximum static friction force is proportional to the normal force and the coefficient of static friction:

* fs = μs * N

**4. Determine the maximum applied force:**

The applied force must be less than or equal to the maximum static friction force to prevent the upper block from slipping:

* F ≤ fs

**5. Substitute and solve:**

From step 3, we know fs = μs * N, and from step 2, we know N = W1. Substituting these into the inequality from step 4:

* F ≤ μs * W1

* F ≤ 0.35 * 4.9 N

* F ≤ 1.715 N

Therefore, the maximum horizontal force that can be applied to the lower block without the upper block slipping is approximately 1.715 N.