Question

For the arrangements shown in figure, acceleration of block B is \[\sqrt 3 \,{\text{m/}}{{\text{s}}^2}\] upwards. Find the normal reaction (n kN) between surfaces of contact of the two blocks.

Answer 1

Hint:Use the expression for Newton’s second law of motion. First determine the acceleration of the block A using a trigonometric equation. Then draw the free body diagram of block A and apply Newton’s second law of motion to block A in horizontal direction. Solve this equation to calculate the value of the normal force.

Formula used:
The expression for Newton’s second law of motion is given by
\[{F_{net}} = ma\] …… (1)
Here, \[{F_{net}}\] is the net force acting on the object, \[m\] is mass of the object and \[a\] is acceleration of the object.

Complete step by step answer:
We have given that the mass of the block A is \[1000\,{\text{kg}}\] and the force F on the block A is \[5000\,{\text{N}}\].
\[{m_A} = 1000\,{\text{kg}}\]
\[\Rightarrow F = 5000\,{\text{N}}\]
We have given that the acceleration of the block B is \[\sqrt 3 \,{\text{m/}}{{\text{s}}^2}\] in the upward direction.
\[{a_y} = \sqrt 3 \,{\text{m/}}{{\text{s}}^2}\]
We have asked to calculate the normal force between the two blocks A and B in contact with each other.

Let us first calculate the acceleration of the block A.The diagram representing these horizontal and vertical accelerations is as follows:

According to trigonometric equations, we can write
\[\cot 30^\circ = \dfrac{{{a_x}}}{{{a_y}}}\]
\[ \Rightarrow {a_x} = {a_y}\cot 30^\circ \]
Substitute \[\sqrt 3 \,{\text{m/}}{{\text{s}}^2}\] for \[{a_y}\] in the above equation.
\[ \Rightarrow {a_x} = \left( {\sqrt 3 \,{\text{m/}}{{\text{s}}^2}} \right)\left( {\sqrt 3 } \right)\]
\[ \Rightarrow {a_x} = 3\,{\text{m/}}{{\text{s}}^2}\]
Hence, the acceleration of the block A is \[3\,{\text{m/}}{{\text{s}}^2}\].

Now we can calculate the normal force between the two blocks.Let us draw the free body diagram of the block A.

In the above free body diagram, \[N\] is the normal force between the surfaces in contact of the two blocks. The horizontal and vertical components of the normal force are shown in the above diagram.

Apply Newton’s second law of motion to the block A in the horizontal direction.
\[F - N\sin 30^\circ = m{a_x}\]

Substitute \[5000\,{\text{N}}\] for \[F\], \[1000\,{\text{kg}}\] for \[m\] and \[3\,{\text{m/}}{{\text{s}}^2}\] for \[{a_x}\] in the above equation.
\[\left( {5000\,{\text{N}}} \right) - N\sin 30^\circ = \left( {1000\,{\text{kg}}} \right)\left( {3\,{\text{m/}}{{\text{s}}^2}} \right)\]
\[ \Rightarrow 5000 - \dfrac{N}{2} = 3000\]
\[ \Rightarrow \dfrac{N}{2} = 5000 - 3000\]
\[ \Rightarrow N = 4000\,{\text{N}}\]
\[ \therefore N = 4\,{\text{kN}}\]

Hence, the normal force between the two surfaces of the blocks which are in contact is \[4\,{\text{kN}}\].

Note: The students may think that the value of acceleration we have calculated is the horizontal component of acceleration of block B. But it is the acceleration of block A. We have used the suffix x with a as it is the acceleration in horizontal direction. Also the students should be careful while determining the angle between the normal force and its components.