Solve the multiple choice questions from the Engineering exam.

Steps to Solve

1. Q1: Calculate the net force

First, sum the forces acting to the right: $5 N + 3 N = 8 N$. Then, find the net force by subtracting the force to the left: $8 N - 10 N = -2 N$. The negative sign indicates the net force is to the left.

2. Q1: Calculate the acceleration

Use Newton's second law, $F = ma$, to find the acceleration: $a = F/m = -2 N / 2 kg = -1 m/s^2$. The negative sign indicates the acceleration is to the left.

3. Q2: Determine the dimensions of force

Force is defined as mass times acceleration ($F=ma$). Mass has dimensions of M, and acceleration has dimensions of $LT^{-2}$. Therefore, the dimensions of force are $MLT^{-2}$.

4. Q3: Identify the pair with non-identical dimensions

Momentum is mass times velocity ($mv$), with dimensions $MLT^{-1}$. Impulse is force times time ($Ft$), with dimensions $(MLT^{-2})T = MLT^{-1}$. So, momentum and impulse have the same dimensions. Torque is force times distance ($Fd$), with dimensions $(MLT^{-2})L = ML^2T^{-2}$. Energy and work also have dimensions $ML^2T^{-2}$. Moment of a force is force times distance ($Fd$), with dimensions $(MLT^{-2})L = ML^2T^{-2}$. Angular momentum is $I\omega$, where $I$ is moment of inertia ($MR^2$) and $\omega$ is angular velocity ($T^{-1}$). Thus, angular momentum has dimensions of $ML^2T^{-1}$.

5. Q4: Define concurrent forces

Concurrent forces are defined as forces whose lines of action meet at a single point.

6. Q5: Determine factors affecting force on a body

The effect of a force depends on its magnitude, its direction, and its line of action (or position).

7. Q6: Determine the unknown variable X

Given $[ML^2T^{-2}] = [X][L]$, we need to find $[X]$. Divide both sides by $[L]$: $[X] = [ML^2T^{-2}] / [L] = [MLT^{-2}]$. Since $[MLT^{-2}]$ represents force, X is likely to represent force.

8. Q7: Calculate the resultant force

Since the forces are perpendicular, use the Pythagorean theorem to find the magnitude of the resultant force: $R = \sqrt{F_1^2 + F_2^2} = \sqrt{(40 N)^2 + (60 N)^2} = \sqrt{1600 + 3600} = \sqrt{5200} \approx 72.11 N$.

9. Q8: Determine that moment of force about O in N-cm

The moment of a force is calculated as the force multiplied by the perpendicular distance from the point of rotation. Given a force $F = 10 N$ acting horizontally at the point $(6, 8)$ cm, the perpendicular distance from the reference point $O$ (0,0) is the y-coordinate, which is $8$ cm. Therefore, the moment $M = F \times d = 10 N \times 8 cm = 80 N \cdot cm$.

10. Q9: Magnitude of resultant of two forces

The magnitude of the resultant of two forces depends on the angle between them

11. Q10: Number of member is truss system

For a simple truss system, the number of members ($m$) is related to the number of joints ($j$) by the equation: $m = 2j - 3$. In this case, $j = 4$, so $m = 2(4) - 3 = 8 - 3 = 5$.