Question

  

Four blocks EACH of mass m = 10 kg are arranged as shown in the picture, on top of a frictionless table. A hand           touching block 1 applies a force of F_h1 = 90 N to the right.  The coefficient of friction between the blocks is           sufficient to keep the blocks from moving with respect to each other.           

What is the total force exerted by block 2 on block 3  ?           

F_23net =

 

Answer 1

Concepts and reason

The concepts used to solve this problem are newton’s second law, and components of force.

Find the acceleration experienced by the system using the force applied on it.

Find the magnitude of the net force acting on the third block using that force’s horizontal and vertical components calculated.

Fundamentals

Newton’s second law states that the acceleration experienced by an object, due to a force that is applied on it, is proportional to the magnitude of that force.

Expression for newton’s second law is,

$F = ma$

Here, $F$ is the force acting on the object, $m$ is the mass of the object, and $a$ is the acceleration produced on the object.

The system has four blocks.

Expression for the mass of the system is,

$m = {m_1} + {m_2} + {m_3} + {m_4}$

Here, ${m_1}$ , ${m_2}$ , ${m_3}$ , and ${m_4}$ are the masses of respectively 1, 2, 3,and 4 blocks.

The entire block has same value of mass.

Force is a vector quantity and hence it can be resolved into it the component form which gives the contribution of this force along the two directions of the coordinate system.

The following figure shows the arrangements of four blocks on top of a frictionless table.

A force $F$ applied on the block 1 act directly on it. For all the blocks to hold the same configuration, this force has to be transferred to all the blocks in the system.

The following figure shows the force exerted by the block 2 on block 3.

Block 2 exerts a force ${F_{23}}$ on the block 3. This force has two components, vertical and horizontal.

The following figure shows the direction of the force exerted by the block 2 on block 3.

The vertical acceleration of the block 3 depends on the vertical component of the force ${F_{23}}$ which is expressed as $F_{23}^y$ .

The horizontal acceleration of the block 3 depends on the horizontal component of the force ${F_{23}}$ which is expressed as $F_{23}^x$ .

Expression for the horizontal component of the force ${F_{23}}$ is

$F_{23}^x = \left( {{m_3} + {m_4}} \right){a_x}$

Here, $F_{23}^x$ is the horizontal component of the force ${F_{23}}$ and ${a_x}$ is the vertical acceleration of the block 3.

Expression for the vertical component of the force ${F_{23}}$ is

$F_{23}^y = \left( {{m_3} + {m_4}} \right){a_y}$

Here, $F_{23}^y$ is the vertical component of the force ${F_{23}}$ and ${a_y}$ is the horizontal acceleration of the block.

The magnitude of the net force is expressed as,

${F_{23}} = \sqrt {{{\left( {F_{23}^x} \right)}^2} + {{\left( {F_{23}^y} \right)}^2}}$ .

Expression for the mass of the system is,

$m = {m_1} + {m_2} + {m_3} + {m_4}$

Substitute $10\,{\rm{kg}}$ for ${m_1}$ , $10\,{\rm{kg}}$ for ${m_2}$ , $10\,{\rm{kg}}$ for ${m_3}$ , and $10\,{\rm{kg}}$ for ${m_4}$ .

$\begin{array}{c}\\m = 10\,{\rm{kg}} + 10\,{\rm{kg}} + 10\,{\rm{kg}} + 10\,{\rm{kg}}\\\\ = 40\,{\rm{kg}}\\\end{array}$

Expression for newton’s second law is,

$F = ma$

Rearrange the above equation to a,

$a = \frac{F}{m}$

Substitute $90\,{\rm{N}}$ for $F$ and $40\,{\rm{kg}}$ for $m$ .

$\begin{array}{c}\\a = \frac{{90\,{\rm{N}}}}{{40\,{\rm{kg}}}}\\\\ = 2.25\,{\rm{m}} \cdot {{\rm{s}}^{ - 2}}\\\end{array}$

Therefore, the acceleration of the system is $2.25\,{\rm{m}}{{\rm{s}}^{ - 2}}$ .

Block two exerts a force ${F_{23}}$ on the block 3. This force can be divided into its components along vertical and horizontal directions.

Expression for the horizontal component of the force is,

$F_{23}^x = \left( {{m_3} + {m_4}} \right){a_x}$

Substitute $10\,{\rm{kg}}$ for ${m_3}$ and ${m_4}$ and $2.25\,{\rm{m}} \cdot {{\rm{s}}^{ - 2}}$ for ${a_x}$ .

$\begin{array}{c}\\F_{23}^x = \left( {20\,{\rm{kg}}} \right)\left( {2.25\,{\rm{m}} \cdot {{\rm{s}}^{ - 2}}} \right)\\\\ = 45\,{\rm{N}}\\\end{array}$

Expression for the vertical component of the force is,

$F_{23}^y = \left( {{m_3} + {m_4}} \right){a_y}$

Substitute $10\,{\rm{kg}}$ for ${m_3}$ and ${m_4}$ and $9.8\,{\rm{m}} \cdot {{\rm{s}}^{ - 2}}$ for ${a_y}$ .

$\begin{array}{c}\\F_{23}^y = \left( {20\,{\rm{kg}}} \right)\left( {9.8\,{\rm{m}} \cdot {{\rm{s}}^{ - 2}}} \right)\\\\ = 196\,{\rm{N}}\\\end{array}$

The net force exerted by the block 2 on block 3 is

${F_{23}} = \sqrt {{{\left( {F_{23}^x} \right)}^2} + {{\left( {F_{23}^y} \right)}^2}}$

Substitute $45\,{\rm{N}}$ for $F_{23}^x$ and $196\,{\rm{N}}$ for $F_{23}^y$ .

$\begin{array}{c}\\{F_{23}} = \sqrt {{{\left( {45\,{\rm{N}}} \right)}^2} + {{\left( {196\,{\rm{N}}} \right)}^2}} \\\\ = 201\,{\rm{N}}\\\end{array}$

Ans:

The net force exerted by the block 2 on block 3 is ${\bf{201}}\,{\bf{N}}$ .