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Q4.7:
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Solution :

Step 1: Given Variables & Initial Setup

The system consists of a body of mass $m = 5 \text{ kg}$ subjected to two mutually perpendicular forces $F_1 = 8 \text{ N}$ and $F_2 = 6 \text{ N}$.

  • Mass ($m$) = $5 \text{ kg}$
  • Force 1 ($F_1$) = $8 \text{ N}$ (Assume along the x-axis)
  • Force 2 ($F_2$) = $6 \text{ N}$ (Assume along the y-axis)

Step 2: Calculating the Resultant Force

Since the forces are perpendicular, the magnitude of the resultant force $F_{net}$ is calculated using the Pythagorean theorem [Per Vector Addition of perpendicular components]:

$F_{net} = \sqrt{F_1^2 + F_2^2}$

$F_{net} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ N}$

Step 3: Calculating Acceleration Magnitude

Applying Newton's Second Law of Motion [$\vec{F} = m\vec{a}$], we determine the acceleration $a$:

$a = \frac{F_{net}}{m}$

$a = \frac{10 \text{ N}}{5 \text{ kg}} = 2 \text{ m/s}^2$

Step 4: Determining Direction

The direction $\theta$ of the acceleration relative to the $8 \text{ N}$ force is given by the trigonometric ratio [Trigonometric Definition of Vector Direction]:

$\tan(\theta) = \frac{F_2}{F_1} = \frac{6}{8} = 0.75$

$\theta = \arctan(0.75) \approx 36.87^\circ$

F1 = 8 N F2 = 6 N F_net = 10 N θ

Step 5: Tabular Summary of Results

Parameter Value
Resultant Force $10 \text{ N}$
Acceleration $2 \text{ m/s}^2$
Direction $36.87^\circ$ from the $8 \text{ N}$ force

Final Solution: The magnitude of the acceleration is $2 \text{ m/s}^2$ directed at an angle of $36.87^\circ$ with respect to the $8 \text{ N}$ force.


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