Class 9 - Unit # 4 :Turning Effect of Forces - Solved Numericals

Physics Numerical Sheet

Unit: Turning Effect of Forces

Forces on Bodies

1. b) A pair of like parallel forces of $15\text{ N}$ each are acting on a body. Find their resultant.
c) Two unlike parallel forces of $10\text{ N}$ each are acting along the same line. Find their resultant.

Data:

Part (b):

• Like Parallel Forces: $F_1 = 15\text{ N}$, $F_2 = 15\text{ N}$
• Resultant Force ($F$) = ?

Part (c):

• Unlike Parallel Forces: $F_1 = 10\text{ N}$, $F_2 = 10\text{ N}$
• Resultant Force ($F$) = ?

Solution:

b) According to the definition of like parallel forces (acting in the same direction):

$$F = F_1 + F_2$$ $$F = 15\text{ N} + 15\text{ N}$$ $$F = 30\text{ N}$$

c) According to the definition of unlike parallel forces acting along the same line (opposing directions):

$$F = F_1 - F_2$$ $$F = 10\text{ N} - 10\text{ N}$$ $$F = 0\text{ N}$$

RESULT: b) The resultant of the like parallel forces is $30\text{ N}$. c) The resultant of the unlike parallel forces is $0\text{ N}$.

Addition of Forces

2. Three forces act on a body: $12\text{ N}$ along the x-axis, $8\text{ N}$ making an angle of $45^\circ$ with the x-axis, and $8\text{ N}$ along the y-axis.
i) Find their resultant magnitude.
ii) Find the direction of the resultant.

Data:

• Force $F_1 = 12\text{ N}$ at $\theta_1 = 0^\circ$
• Force $F_2 = 8\text{ N}$ at $\theta_2 = 45^\circ$
• Force $F_3 = 8\text{ N}$ at $\theta_3 = 90^\circ$
• Resultant Force ($F$) = ?
• Direction ($\theta$) = ?

Solution:

First, resolve each force into its respective horizontal ($x$) and vertical ($y$) components:

For x-components ($F_x = F \cos\theta$):

$$F_{1x} = 12 \cos(0^\circ) = 12 \times 1 = 12\text{ N}$$ $$F_{2x} = 8 \cos(45^\circ) = 8 \times 0.707 = 5.66\text{ N}$$ $$F_{3x} = 8 \cos(90^\circ) = 8 \times 0 = 0\text{ N}$$

For y-components ($F_y = F \sin\theta$):

$$F_{1y} = 12 \sin(0^\circ) = 12 \times 0 = 0\text{ N}$$ $$F_{2y} = 8 \sin(45^\circ) = 8 \times 0.707 = 5.66\text{ N}$$ $$F_{3y} = 8 \sin(90^\circ) = 8 \times 1 = 8\text{ N}$$

Net x-component ($F_x$) and Net y-component ($F_y$) of the resultant:

$$F_x = F_{1x} + F_{2x} + F_{3x} = 12 + 5.66 + 0 = 17.66\text{ N}$$ $$F_y = F_{1y} + F_{2y} + F_{3y} = 0 + 5.66 + 8 = 13.66\text{ N}$$

i) Magnitude of Resultant Force ($F$):

$$F = \sqrt{F_x^2 + F_y^2}$$ $$F = \sqrt{(17.66)^2 + (13.66)^2}$$ $$F = \sqrt{311.88 + 186.59} = \sqrt{498.47} \approx 22.33\text{ N}$$

ii) Direction of the Resultant Force ($\theta$):

$$\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)$$ $$\theta = \tan^{-1}\left(\frac{13.66}{17.66}\right) = \tan^{-1}(0.7735) \approx 37.7^\circ$$

RESULT: The magnitude of the resultant force is approximately $22.3\text{ N}$ (or $22.2\text{ N}$ rounded) and its direction is $37.7^\circ$ with the x-axis.

Resolution of Forces

3. b) A gardener is driving a lawnmower with a force of $80\text{ N}$ that makes an angle of $40^\circ$ with the ground.
i) Find its horizontal component.
ii) Find its vertical component.

Data:

• Applied Force ($F$) = $80\text{ N}$
• Angle ($\theta$) = $40^\circ$
• Horizontal component ($F_x$) = ?
• Vertical component ($F_y$) = ?

Solution:

i) For Horizontal Component ($F_x$):

$$F_x = F \cos\theta$$ $$F_x = 80 \times \cos(40^\circ)$$ $$F_x = 80 \times 0.7660$$ $$F_x = 61.28\text{ N}$$

ii) For Vertical Component ($F_y$):

$$F_y = F \sin\theta$$ $$F_y = 80 \times \sin(40^\circ)$$ $$F_y = 80 \times 0.6428$$ $$F_y = 51.42\text{ N}$$

RESULT: The horizontal component of the force is $61.28\text{ N}$ and the vertical component is $51.42\text{ N}$.

4. b) The horizontal and vertical components of a force are $4\text{ N}$ and $3\text{ N}$ respectively. Find:
i) The resultant force.
ii) The direction of the resultant.

Data:

• Horizontal component ($F_x$) = $4\text{ N}$
• Vertical component ($F_y$) = $3\text{ N}$
• Resultant Force ($F$) = ?
• Direction ($\theta$) = ?

Solution:

i) Resultant Force ($F$) is given by:

$$F = \sqrt{F_x^2 + F_y^2}$$ $$F = \sqrt{4^2 + 3^2}$$ $$F = \sqrt{16 + 9} = \sqrt{25} = 5\text{ N}$$

ii) Direction of the resultant force ($\theta$) is given by:

$$\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)$$ $$\theta = \tan^{-1}\left(\frac{3}{4}\right) = \tan^{-1}(0.75) \approx 36.87^\circ$$

RESULT: The magnitude of the resultant force is $5\text{ N}$ and its direction is $36.87^\circ$.

Moment of Force (Torque)

5. b) A spanner of $0.3\text{ m}$ length can produce a torque of $300\text{ Nm}$.
i) Determine the force applied on it.
ii) What should be the length of the spanner if the torque is to be increased to $500\text{ Nm}$ with the same applied force?

Data:

Part (i):

• Length / Moment Arm ($d_1$) = $0.3\text{ m}$
• Torque ($\tau_1$) = $300\text{ Nm}$
• Force ($F$) = ?

Part (ii):

• New Torque ($\tau_2$) = $500\text{ Nm}$
• Force ($F$) = same as calculated in part i
• New Length ($d_2$) = ?

Solution:

(i) According to the definition of Torque ($\tau = F \times d$):

$$F = \frac{\tau_1}{d_1}$$ $$F = \frac{300\text{ Nm}}{0.3\text{ m}} = 1000\text{ N}$$

(ii) Keeping the force constant ($F = 1000\text{ N}$), find the new length ($d_2$):

$$\tau_2 = F \times d_2 \implies d_2 = \frac{\tau_2}{F}$$ $$d_2 = \frac{500\text{ Nm}}{1000\text{ N}} = 0.5\text{ m}$$

RESULT: (i) The force applied is $1000\text{ N}$. (ii) The required length of the spanner is $0.5\text{ m}$.

Principle of Moments

6. b) A uniform meter rule supported at its center is balanced by two forces, $12\text{ N}$ and $20\text{ N}$.
i) If the $20\text{ N}$ force is placed at a distance of $3\text{ m}$ from the pivot, find the position of the $12\text{ N}$ force on the other side of the pivot.
ii) If the $20\text{ N}$ force is moved to $4\text{ m}$ from the pivot, find the force needed to replace the $12\text{ N}$ force to maintain balance.

Data:

• First Force ($F_1$) = $12\text{ N}$
• Second Force ($F_2$) = $20\text{ N}$
• (i) Distance of 2nd Force ($d_2$) = $3\text{ m}$, Distance of 1st Force ($d_1$) = ?
• (ii) New Distance of 2nd Force ($d'_2$) = $4\text{ m}$, New First Force ($F'_1$) = ? at distance $d_1$

Solution:

(i) According to the Principle of Moments ($\text{Clockwise Moments} = \text{Anticlockwise Moments}$):

$$F_1 \times d_1 = F_2 \times d_2$$ $$12\text{ N} \times d_1 = 20\text{ N} \times 3\text{ m}$$ $$d_1 = \frac{60}{12} = 5\text{ m}$$

(ii) If the $20\text{ N}$ force is moved to $4\text{ m}$ ($d'_2 = 4\text{ m}$), keeping the position $d_1 = 5\text{ m}$ fixed:

$$F'_1 \times d_1 = F_2 \times d'_2$$ $$F'_1 \times 5\text{ m} = 20\text{ N} \times 4\text{ m}$$ $$F'_1 = \frac{80}{5} = 16\text{ N}$$

RESULT: (i) The position of the $12\text{ N}$ force is $5\text{ m}$ from the pivot. (ii) The replacement force required is $16\text{ N}$.

Couple

8. b) A mechanic uses a double-arm spanner to turn a nut. He applies a force of $15\text{ N}$ at each end of the spanner and produces a torque of $60\text{ Nm}$. What is the length of the moment arm of the couple?
c) If he wants to produce a torque of $80\text{ Nm}$ with the same spanner, how much force must he apply?

Data:

Part (b):

• Force of the couple ($F$) = $15\text{ N}$
• Torque produced ($\tau$) = $60\text{ Nm}$
• Arm of the couple ($d$) = ?

Part (c):

• New Torque ($\tau'$) = $80\text{ Nm}$
• Arm of the couple ($d$) = calculated in part b
• New Force required ($F'$) = ?

Solution:

(b) According to the definition of a Couple ($\tau = F \times d$):

$$d = \frac{\tau}{F}$$ $$d = \frac{60\text{ Nm}}{15\text{ N}} = 4\text{ m}$$

(c) To produce a torque of $80\text{ Nm}$ using the same arm length ($d = 4\text{ m}$):

$$\tau' = F' \times d \implies F' = \frac{\tau'}{d}$$ $$F' = \frac{80\text{ Nm}}{4\text{ m}} = 20\text{ N}$$

RESULT: (b) The length of the moment arm of the couple is $4\text{ m}$. (c) The new force required is $20\text{ N}$.

Equilibrium

9. A uniform metre rule is balanced at the $30\text{ cm}$ mark when a load of $0.80\text{ N}$ is hung at the zero ($0\text{ cm}$) mark.
i) At what point on the rule is the center of gravity located?
ii) Calculate the weight of the rule.

Data:

• Pivot position = $30\text{ cm}$
• Hanging load ($W_1$) = $0.80\text{ N}$ at position $0\text{ cm}$
• Moment arm of load ($d_1$) = $30\text{ cm} - 0\text{ cm} = 30\text{ cm} = 0.3\text{ m}$
• Center of Gravity ($C$) = ?
• Weight of rule ($W_2$) = ?

Solution:

(i) Position of Center of Gravity:

Since the metre rule is specified to be uniform, its geometric balance point and center of gravity lies exactly at its midpoint.

$$\text{Center of Gravity} = 50\text{ cm mark}$$

(ii) Calculate the weight of the rule ($W_2$):

The weight of the rule acts downwards directly at its center of gravity ($50\text{ cm}$).
The distance from the pivot ($30\text{ cm}$) to the center of gravity ($50\text{ cm}$) forms the moment arm ($d_2$):

$$d_2 = 50\text{ cm} - 30\text{ cm} = 20\text{ cm} = 0.2\text{ m}$$

Applying the Second Condition of Equilibrium ($\sum \tau = 0$):

$$\text{Anticlockwise Moment} = \text{Clockwise Moment}$$ $$W_1 \times d_1 = W_2 \times d_2$$ $$0.80\text{ N} \times 0.3\text{ m} = W_2 \times 0.2\text{ m}$$ $$0.24 = W_2 \times 0.2$$ $$W_2 = \frac{0.24}{0.2} = 1.2\text{ N}$$

RESULT: i) The center of gravity of the rule is at the $50\text{ cm}$ mark. ii) The total weight of the metre rule is $1.2\text{ N}$.