We discuss problems, which highlight certain aspects of the study leading to friction. The questions are categorized in terms of the characterizing features of the subject matter :

Problem 4 : A variable increasing force is applied on a rod in the arrangement shown in the figure. The blocks are identical and are placed over a rough horizontal surface. Determine which of the block will move first?

Figure 7: A variable increasing force is applied on a rod.

Two blocks and a rod

Solution : The blocks undergo translation in opposite directions. Since blocks are identical, the friction between block and horizontal surface is same for both blocks. It is evident that the block, having greater external force parallel to surface, will overcome friction first.

It is, therefore, evident that we need to analyze forces in the horizontal direction. The forces on the rod are shown in the figure below. The components of forces in horizontal direction form balanced force system before the motion is initiated. From the force analysis in horizontal direction, we have :

Figure 8: Forces on a rod.

Two blocks and a rod

⇒ N 2 sin θ = N 1 + F ⇒ N 2 sin θ = N 1 + F

The contact normal force are equal and opposite between rod and blocks. The horizontal component of normal force ( N 2 sin θ N 2 sinθ ) as applied by the rod on the block “B” is greater than the horizontal normal force, ( N 1 N 1 ) as applied by the rod on the block “A”. It means the external horizontal force on “B” will exceed the limiting friction first. As such, block “B” will move first.

Problem 5 : The masses of three blocks stacked over one another are m 1 m 1 = 30 kg, m 2 m 2 = 10 kg and m 3 m 3 = 40 kg. The coefficients of friction between block “1” and ground is μ 0 μ 0 = 0.1, between “1” and “2” is μ 1 μ 1 = 0.2 and between “2” and “3” is μ 2 μ 2 = 0.3. An external force is applied on the block “3” as shown in the figure. What least force is required to initiate motion of any part of the stack?

Figure 9: Blocks of different masses are placed one over other.

Three stacked blocks

Solution : In order to answer this question, we are required to determine limiting static friction at each of the interface. The limiting friction between first block and ground is :

F 0 = μ 0 N = μ 0 m 1 + m 2 + m 3 g = 0.1 30 + 10 + 40 X 10 = 80 N F 0 = μ 0 N = μ 0 m 1 + m 2 + m 3 g = 0.1 30 + 10 + 40 X 10 = 80 N

The limiting friction between first and second block is :

F 1 = μ 1 N = μ 1 m 2 + m 3 g = 0.2 10 + 40 X 10 = 100 N F 1 = μ 1 N = μ 1 m 2 + m 3 g = 0.2 10 + 40 X 10 = 100 N

The limiting friction between second and third is :

F 2 = μ 2 N = μ 2 m 3 g = 0.3 X 40 X 10 = 120 N F 2 = μ 2 N = μ 2 m 3 g = 0.3 X 40 X 10 = 120 N

If the external force is less than 80 N, then none of the block will move. If external force is equal or greater than 80 N, but less than 100 N, then only the first block (the one in contact with ground) will move as external force on it is equal to the limiting friction between block “1” and the underlying horizontal surface.