3.1 Define the term normal force - NSC Technical Sciences - Question 3 - 2021 - Paper 1

To draw the free-body diagram for block A, identify the following forces:

1. The applied force ($F_A = 50 ext{ N}$) acting horizontally to the right.
2. The kinetic friction force ($F_{fA}$) acting to the left, opposing the motion, which can be calculated as $F_{fA} = ext{μ}_k imes N_A$ where $N_A$ is the normal force on block A.
3. The gravitational force ($F_{gA} = m_A imes g = 25 ext{ kg} imes 9.8 ext{ m/s}^2 = 245 ext{ N}$) acting downwards.
4. The normal force ($N_A$) acting upwards, which balances the gravitational force.

Make sure to label each force with appropriate arrows indicating direction.

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In equation form, it is expressed as $F_{net} = m imes a$, where $F_{net}$ is the net force, $m$ is the mass, and $a$ is the acceleration.

To calculate the coefficient of kinetic friction ($ext{μ}_k$), use the formula:

$ext{μ}_k = \frac{F_{fA}}{N_A}$

From the free-body diagram, we need to find $N_A$:

1. The normal force can be determined as: $N_A = F_{gA} = m_A imes g = 25 ext{ kg} imes 9.8 ext{ m/s}^2 = 245 ext{ N}$

2. The frictional force can be described as: $F_{fA} = F_A - m_A imes a$
   For block A:

• With $F_A = 50 ext{ N}$ and calculated values substituted, we find:

3. To find the value of $ext{μ}_k$: $ext{μ}_k = \frac{F_{fA}}{N_A} = \frac{5.82}{221.53} ext{ N} = 0.026$

Using free-body analysis for Block B:

1. Set net force and calculate: $F_{net} = F_B - T - F_{fB}$

2. Rearranging gives: $T = F_B - F_{net}$

3. Substitute the values:

• Knowing the forces: $F_{B} = 350 ext{ N}$ and friction force. Calculate the net force: $F_{net} = m_B imes a$

4. Hence, tension will yield to: $T = 350 ext{ N} - 45 ext{ kg} imes 9.8 ext{ m/s}^2 = 154.14 ext{ N}$