Question

A steel block ,S, is placed on the concrete platform, B. The weight of the block is 5N. A steadily increasing force is applied horizontally to the block so that it just begins to move . If the coefficient of friction is 0.2. Calculate the minimum force required to move the block.

Answer 1

To find the minimum force required to move the steel block, we can use the concept of friction. The force of static friction, which must be overcome to get the block moving, can be calculated using the equation:

\[
f_{s} = \mu_s \cdot N
\]

where:
- \( f_{s} \) is the force of static friction,
- \( \mu_s \) is the coefficient of static friction (given as 0.2), and
- \( N \) is the normal force.

Since the block is on a horizontal surface and the only vertical force acting on it (besides any potential vertical force from the applied force) is its weight, the normal force \( N \) is equal to the weight of the block:

\[
N = \text{weight of the block} = 5 \, \text{N}
\]

Now we can calculate the force of static friction:

\[
f_{s} = \mu_s \cdot N = 0.2 \cdot 5 \, \text{N} = 1 \, \text{N}
\]

Thus, the minimum force required to move the block is equal to the force of static friction, which is:

\[
\text{Minimum force required} = 1 \, \text{N}
\]

Therefore, the minimum force required to move the block is **1 N**.