Question

A 20 N horizontal force F pushes a block weighing 3.0 N against a vertical wall. The coefficient of static friction between the wall and the block is 0.60, and the coefficient of kinetic friction is 0.40. Assume that the block is not moving initially.


In unit-vector notation, what is the force exerted on the block by the wall?
(___)N ihad + (____)N jhad

A 20 N horizontal force F pushes a block weighing 3.0 N against a vertical wall. The coefficient of static friction between the wall and the block is 0.60, and the coefficient of kinetic friction is 0.40. Assume that the block is not moving initially. In unit-vector notation, what is the force exerted on the block by the wall? (___)N ihad + (____)N jhad

Answer 1

Concepts and reason

The concepts used to solve this problem are frictional force, static equilibrium condition, vector algebra.

Draw the free body diagram of the system and apply the equilibrium condition to the force exerted by the wall.

Calculate the vertical and horizontal frictional force. After that calculate the total force exerted by wall on the block.

Fundamentals

The expression for weight $\left( W \right)$ of a body is:

$W = mg$

Here $m$ is the mass and $g$ is acceleration due to gravity.

Write the expression for frictional force $\left( f \right)$ .

$f = {\mu _s}{F_N}$

Here ${\mu _s}$ is the coefficient of static friction and ${F_N}$ is the normal reaction force acting on the body.

Write the expression for force in vector notation.

$F = {\left( {{F_{x}}} \right)_{net}}\hat i + {\left( {{F_y}} \right)_{net}}\hat j$

Here, ${\left( {{F_x}} \right)_{net}}$ is the net force acting on the body in x direction and ${\left( {{F_y}} \right)_{net}}$ is the net force acting on the body in y direction.

Draw free body diagram of the system.

Here, $f$ is the frictional force between the wall and the block , $m$ is the mass of the block, ${F_N}$ is the normal reaction force acting on the block and $g$ is the acceleration due to gravity.

From the free body diagram, the static equilibrium condition for vertical and horizontal forces.

The horizontal force is,

$\begin{array}{l}\\F - {F_N} = 0\\\\F = {F_N}\\\end{array}$

Substitute $20{\rm{ N}}$ for $F$ .

${F_N} = 20{\rm{ N}}$

The vertical force is,

$\begin{array}{l}\\f - W = 0\\\\f = W\\\end{array}$

Substitute $3{\rm{ N}}$ for $W$ .

$f = 3{\rm{ N}}$

Write the expression for frictional force.

${f_s} = {\mu _s}{F_N}$

Here ${\mu _s}$ is the coefficient of static friction.

Substitute $0.60$ for ${\mu _s}$ and $20{\rm{ N}}$ for ${F_N}$ .

$\begin{array}{c}\\{f_s} = \left( {0.60} \right)\left( {20{\rm{ N}}} \right)\\\\ = 12{\rm{ N}}\\\end{array}$

The frictional force $f$ acting on the object. Now if $f < {\mu _s}{F_N}$ then the block will not slide and if $f > {\mu _s}{F_N}$ the block will slide.

Here,

$3{\rm{ N < 12 N}}$

Thus, the block does not move.

Write the expression for net force in the horizontal direction.

${\left( {{F_x}} \right)_{net}} = - {F_N}$

The negative sign is taken as ${F_N}$ acts along the negative x axis.

Substitute 20 N for ${F_N}$ .

${\left( {{F_x}} \right)_{net}} = - 20{\rm{ N}}$

Write the expression for net force in the vertical direction.

${\left( {{F_y}} \right)_{net}} = f$

Substitute $3{\rm{ N}}$ for $f$ .

${\left( {{F_y}} \right)_{net}} = 3{\rm{ N}}$

Write the expression for the force exerted by the wall on the block in vector form.

$F = {\left( {{F_{x}}} \right)_{net}}\widehat i + {\left( {{F_y}} \right)_{net}}\widehat j$

Substitute $- 20{\rm{ N}}$ for ${\left( {{F_x}} \right)_{net}}$ and $3{\rm{ N}}$ for ${\left( {{F_y}} \right)_{net}}$ .

$F = - 20{\rm{ N }}\widehat i + 3{\rm{ N }}\widehat j$

Ans:

The force exerted by the wall on the block is $F = - 20{\rm{ N }}\widehat i + 3{\rm{ N }}\widehat j$ .