Question

A chair of weight 95.0N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 39.0N directed at an angle of 37.0 degrees below the horizontal and the chair slides along the floor. 1)Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.

Answer 1

Concepts and reason

The concepts that are required to solve this problem are Newton’s second law of motion, equilibrium of forces, resolution of forces, and normal reaction.

Free body diagram is drawn to show all the force acting on the body. The condition for the equilibrium of forces along the vertical direction is to be applied and then equations are obtained to determine the magnitude of the normal reaction.

Fundamentals

The Newton’s second law of motion states that the rate of change of momentum is called the force. Mathematically,

$F = \frac{d}{{dt}}\left( p \right)$

Here, $p$ is the is the momentum,

Write the expression of momentum,

$p = mv$

Here, $m$ is the mass of the body, and $v$ is the velocity.

Substitute, $mv$ for $p$ in the expression for Newton’s second law of motion .

$\begin{array}{c}\\F = \frac{d}{{dt}}\left( {mv} \right)\\\\ = m\frac{{dv}}{{dt}}\\\\ = ma\\\end{array}$

Here, $F$ is the force, and $a$ is the acceleration.

The expression of the weight of an object is,

$W = mg$

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

Equilibrium of Forces:

If two or more forces which are acting on the body. The resultant of all forces is zero then this condition is called as equilibrium of forces.

Resolution of forces:

In two dimensional axes, any force at any direction contain a position from x and y axis. Hence, this force can be resolve in two components that is horizontal or x-component and vertical or y-component.

Normal reaction: It is the reaction force of the surface on the body, on which the body is placed upon.

Write the conditions for the equilibrium of forces.

$\begin{array}{c}\\\Sigma {F_x} = 0\\\\\Sigma {F_y} = 0\\\end{array}$

Here, $\Sigma {F_x}$ is the sum of all the forces along the horizontal direction, and $\Sigma {F_y}$ is the sum of all the forces along the vertical direction.

For a force $V$ making an angle $\theta$ with the positive x-axis, the force resolution to obtain the rectangular components is shown as,

The horizontal component $\left( {{V_x}} \right)$ and vertical component $\left( {{V_y}} \right)$ is,

${V_x} = V\cos \theta$

${V_y} = V\sin \theta$

(1)

Draw the diagram of the provided arrangement.

Resolve the force along its components and draw a normal reaction upwards perpendicular to the ground. Draw the free body diagram.

Consider the free body diagram of the chair and apply the condition of equilibrium along the vertical direction.

$\begin{array}{l}\\\Sigma {F_y} = 0\\\\N - W - F\sin \theta = 0\\\\N = W + F\sin \theta \\\end{array}$

Substitute $90{\rm{ N}}$ for $W$ , $39{\rm{ N}}$ for $F$ and $37^\circ$ for $\theta$ in the above expression.

$\begin{array}{c}\\N = 95.0{\rm{ N}} + 39.0{\rm{ N}}\left( {\sin 37.0^\circ } \right)\\\\ = 118.47{\rm{ N}}\\\\ \simeq {\rm{118}}{\rm{.5}}\;{\rm{N}}\\\end{array}$

Ans: Part 1

The magnitude of the normal reaction is ${\rm{118}}{\rm{.5}}\;{\rm{N}}$ .