Turning Effect Of Forces - Physics For Class IX (Science Group) - Numericals

Go To Index

Physics For Class IX (Science Group)
UNIT 4: TURNING EFFECT OF FORCES
Numericals

Worked Example 1

1. Find the resultant of three forces 15 N along x-axis, 10 N making an angle of 30° with x- axis and 10 N along y-axis.
Solution:
Step 1: Write the Known quantities and choose a suitable scale.
Here,

• F₁ = 15 N along x-axis
• F₂ = 10 N 30° with x-axis
• F₃ = 10 N along Y-axis.
• Scale 2 N = 1 cm.

Step 2: Draw the representative vectors for the forces F₁, F₂, F₃ according to the scale in the given directions.

Step 3: Take F₁ as first vector and draw F₂ and F₃ in such away that the tail of next vector coincides with the headof the previous vector.

Step 4: Join the tail of the F₁ with the head of F₃ with a straight-line F with an arrow pointing towards F₃.
According to head to tail rule, Force F represents the resultant force.

Step 5: Measure the length of F with a ruler and multiply it with 2 Ncm^-1 that is the magnitude of resultant. Measure the angle with protector that F makes with F₁. This gives the direction of resultant Force.

Worked Example 2

2. A man is pushing a wheelbarrow on a horizontal ground with a force of 300 N making an angle of 60° with ground. Find the horizontal and vertical components of the force.
Solution:
Step 1: Write the known quantities and point out the quantities to be found.

• F = 200 N
• θ = 60° with horizontal.
• F_x = ?
• F_y =?

Step 2: Write the formula and rearrange if necessary.

• F_x = F cos θ
• F_y = F sin θ

Step 3: Put the values in the formula and calculate.

F_x = F cos 60°
F_x = 300 N x cos 60°
F_x = 300 N x 0.5
F_x = 150 N

F_y = F sin 60°
F_y = 300 N x sin 60°
F_y = 300 N x 0.8660
F_y = 259.8N

Ans: Therefore, horizontal and vertical components of pushing force are 150 N and 259.8 N respectively.

Worked Example 3

3. A car driver tightens the nut of wheel using 20 cm long spanner by exerting a force of 300N. Find the torque.
Solution:
Step 1: Write the known quantities and point out the quantity to be found.

• F = 300 N
• L = 20 cm = 0.20 m
• τ = ?

Step 2: Write the formula and rearrange if necessary.

τ = F × L

Step 3: Put the values in formula and calculate

τ = 300 N × 0.20 m = 60 Nm

Ans: Thus, torque of 60 Nm is used to tighten the nut.

Worked Example 4

4. Consider a meter rod supported at mid-point O as shown in figure. The block of 20 N is suspended at point A 30 cm from O. Find the weight of the block that balances it at point B, 20 cm from O.
Solution:
Step 1: write known quantities and point out unknown quantities.

• W₁ = 20 N
• Moment arm of W₁ = OA = 30 cm = 0.30 m
• Moment arm of W₂ = OB = 20 cm
• W₂ = ?

Worked Example 5

5. A uniform rod of length 2.0 m is placed on a wedge at 0.5 m from its one end. A force of 150 N is applied at one of its ends near the wedge to keep it horizontal. Find the weight of the rod and the reaction of the wedge.
Solution:
Step 1: Write the known quantities and point out unknown.

• F = 150 N
• OA = 0.5 m
• AG = BG = 1.0 m
• OG =AG – AO =1.0 m - 0.5 m = 0.5 m
• W = ?
• R = ?

Step 2: Write formula and substitute values. For W applying second condition of equilibrium, taking torques about O.

Σ𝜏 = 0

F x AO + R x 0 + W x OG = 0

150 x 0.5 - W x 0.5 = 0 or

W x 0.5 m = 150 x 0.5 m

Numericals (Self Assessment Question)

Q.8: What will be moment of force? When 500 N force is applied on a 40 cm long spanner to tighten a nut.
Solution:
Data:

• F = 500 N
• L = 40 cm = 40/ 100 = 0.40 m
• τ = ?

Formula:

τ = F × L

Calculation:

τ = 500 × 0.40 =200 Nm.

Ans. Thus, the moment of force will be 200 Nm.

Q.12: If two forces 5 N each form a couple and the moment arm is 0.5 m .Then what will be torque of the couple?
Solution:
Data:

• F = 5 N
• d = 0.5 cm
• τ = ?

Formula:

τ = F x d

Calculation:

τ = 50 x 0.5 = 2.5 Nm

Ans: Thus, The torque of the couple will be 2.5 Nm.

Numericals (Text Book exercise)

Section (B) Structured Questions
Forces on bodies:
1. b) A pair of like parallel forces 15 N each are acting on a body. Find their resultant.
Ans: SOLUTION:
Data:

• F₁ = 15 N
• F₂ = 15 N
• F = ?

Formula:

F = F₁ + F₂

Calculation:

F = 15 + 15 = 30 N

Ans. Their resultant force is 30 N.

1. c) Two unlike parallel forces 10 N each acting along same line. Find their resultant.
Ans: SOLUTION:
Data:

• F₁ = 10 N
• F₂ = 10 N
• F = ?

Formula:

F = F₁ - F₂

Calculation:

F = 10 - 10 = 0 N

Ans. Their resultant force is Zero.

Addition of forces
2. b) Three forces 12 N along x-axis, 8 N making an angle of 45° with x-axis and 8 N along y-axis.
i) Find their resultant
ii) Find the direction of resultant
Ans: SOLUTION:
Data:

• F₁ = 12 N along x-axis
• F₂ = 8 N making an angle of 45° with x-axis
• F₃ = 8 N along y-axis

Graphical Representation:
Scale: 1 N = 1 square (You can also take 2 N = 1 square)

Ans:

1. Magnitude of Resultant of Force: Measuring the length of F with ruler, it is found that force is 22.322 N.
2. Direction of resultant: Measuring the angle with the protector (Dee) that F makes with F₁, it is found 37.72°

Resolution of forces
3. b) A gardener is driving a lawnmower with a force o of 80 N that makes an angle of 40° with the ground.
i) Find its horizontal component
ii) Find its vertical component
Ans: Solution
Data:

• Force F = 80 N
• Angle = θ = 40°
• Horizontal Component = F_x = ?
• Vertical Component = F_y = ?

Formula:

1. F_x = F cos θ
2. F_y = F sin θ

Calculation:

F_x = F cos θ

F_x = (80) cos 40°

F_x = (80) x (0.766)

F_x = 61.28 N

F_y = F sin θ

F_y = (80) sin 40°

F_y = (80) x (0.642)

F_y = 51.36 N

Ans: Therefore, Horizontal and vertical components of the force are 61.28 N and 51.36 N respectively.

4. b) Horizontal and vertical components of a force are 4 N and 3 N respectively. Find
i) Resultant force
ii) Direction of resultant
Ans: Solution
Data:

• Horizontal Component = F_x = 4 N
• Vertical Component = F_y = 3 N
• Resultant Force = F = ?
• Direction of Resultant = θ

Moment of force
5. b) A spanner of 0.3 m length can produce a torque of 300 Nm.
i) determine the force applied on it.
ii) What should be the length of the spanner if torque is to be increased to 500 Nm with same applied force.
Ans: Solution:
(i) Applied Force
Data:

• Length of the Spanner = L = 0.3 m
• Torque = 𝜏 = 300 Nm
• Applied Force = F = ?

Formula:

𝜏 = F x L

Calculation:

300 = F x 0.3 or

Applied force = F = 1000 N

(ii) Length of spanner
Data:

• Applied Force = F = 1000 N
• Torque = 𝜏 = 500 Nm
• Length of the Spanner = L = ?

Formula:

𝜏 = F x L

Calculation:

500 = 1000 x L or

Length of Spanner = L = 0.5 m

Ans: Therefore,

1. The applied force on the spanner is 1000 N. and
2. The length of the spanner will be 0.5 m.

Principle of moments
6. b) A uniform meter rule is supported at its center is balanced by two forces 12 N and 20 N
i) if 20 N force is placed at a distance of 3 m from pivot find the position of 12 N force on the other side of pivot
ii) if the 20 N force is moved to 4 cm from pivot then find force to replace 12 N force.
Ans: Solution:
(i) Position Of Force
Data:

• 1^st force = F₁ = 20 N
• 2^nd force = F₂ = 12 N
• Moment of arm of F₁ = d₁ = 3 m
• Moment of arm of F₂ = d₂ = ?

Formula:

Clockwise Moment = Anticlockwise Moment

F₁ x d₁ = F₂ x d₂

Calculation:

20 x 3 = 12 x d₂

60 = 12 x d₂

60 = 12d₂

Ans: Therefore, The position or the moment arm of force F₂ on the other side of the pivot is 5 m.

(i) 2^nd Force
Data:

• 1^st force = F₁ = 20 N
• 2^nd force = F₂ = ?
• Moment of arm of F₁ = d₁ = 4 m
• Moment of arm of F₂ = d₂ = 5 m

Formula:

Clockwise Moment = Anticlockwise Moment

F₁ x d₁ = F₂ x d₂

Calculation:

20 x 4 = F₂ x 5

80 = F₂ x 5

80 = 5F₂

Ans: Therefore, if the 20 N force is moved to 4 cm from pivot then the 12 N force would be replaced by 16 N force.

Couple
8. b) A mechanic uses a double arm spanner to turn a nut. He applies a force of 15 N at each end of the spanner and produces a torque of 60 Nm. What is the length of the moment arm of the couple?
Ans: Solution:
Data:

• Applied Force = F = 15 N
• Torque produced = 𝜏 = 60 Nm
• Length of the moment arm of the couple = d = ?

Formula:

𝜏 = F x d

Calculation:

60 = 15 x d

d =   ⁶⁰/₁₅  = 4 m

Ans: Therefore, the length of the moment arm of the couple is 4 m.

8. c) If he wants to produce a torque of 80 Nm with same spanner then how much force he should apply?
Ans: Solution:
Data:

• Applied Force = F = ?
• Torque produced = 𝜏 = 80 Nm
• Length of the moment arm of the couple = d = 4 m (Calculated in Q.8. b above)

Formula:

𝜏 = F x d

Calculation:

80 = F x 4

F =  ⁸⁰/₄   = 20 N

Ans: Therefore, If he wants to produce a torque of 80 Nm with same spanner then he should apply 20 N force.

Equilibrium
9. b) A uniform metre rule is balanced at the 30 cm mark when a load of 0.80 N is hung at the zero mark.
i) At what point on the rule is the Centre of gravity of the rule?
ii) calculate the weight of the rule?
Ans: Solution:
(i) Data:
Load = 0.80 N

• Moment arm of F = OA = 30 cm = ³⁰/₁₀₀ = 0.3 m
• Length of the rule = L = 1 m
• The Centre of gravity of the rule = ?

(ii) Data:

• Load = 0.80 N
• Moment arm of F = OA = 30 cm = ³⁰/₁₀₀ = 0.3 m
• Weight of the rule = W = ?
• Moment arm of W = OG = 20 cm = ²⁰/₁₀₀ = 0.20 m

Formula:

∑𝜏 = 0

Calculation:

F x OA + R x 0 - W x OG = 0

0.8 x 0.3 + 0 - W x 0.20 = 0

0.24 - W x 0.20 = 0

W x 0.20 = 0.24

W =   ^0.24/_0.20  = 1.2 N

Ans: Therefore, The weight of the rule is 1.2 N.

More Numericals

10) A telephone pole of mass 300 kg is 30 m long. Its center of gravity is 10 m from the thick end. What force must be applied at the thin end to maintain the pole in horizontal position when it is supported at its mid point?
Ans: Solution:
Data:

• Mass of the pole = 300 kg
• Length of the pole = 30 m
• Center of gravity = 10 m
• Weight of the pole = W = mg = 300 x 9.8 = 2940 N

Required Data:

• Force = F = ?

Formula:

∑𝜏 = 0

Calculation:
Forces acting on the pole are shown in the figure where W represents the weight of the pole and F represents the force applied to the thin end to maintain the pole in a horizontal position when it is supported at its mid point.
From figure we have,

• AB = 30 m,
• AC = 15 m,
• BC = 15 m,
• DC = 5m

Now, According to 2nd condition of equilibrium about C, we have

∑𝜏 = 0

W x DC - F x BC = 0

2940 x 5 - F x 15 = 0

14700 - 15F = 0

15F = 14700

F =  ¹⁴⁷⁰⁰/₁₅

F = 980 N

Ans: Therefore, To maintain the pole in horizontal position 980 N force must be applied at the thin end.

11) A uniform rod 10 meters long and weighing 30 N is supported in a horizontal position on a fulcrum with weight of 40 N and 50 N suspended from its ends as shown in the figure. Find the position of the fulcrum.
Ans: Solution:
Data:

• The length of rod = AB = 10 m
• The weight of point C = W_c = 30 N
• The weight of point A = W_a = 4 0N
• The weight of point B = W_g = 50 N
• The length of rod between A and C = 5 m

Required Data:

• Position of fulcrum = x = ?

Formula:
(i) 1^st condition of equilibrium:
F = F_x + F_y Or
F - F_x - F_y = 0
(ii) According to 2^nd condition of Equilibrium
∑𝜏 = 0

Calculation:
The weight of the uniform rod acts at “C”. Let D be the fulcrum and F be the upward force at the fulcrum. Suppose x is the position of the fulcrum with respect to A. The rod is in equilibrium.
(i) Now by applying 1^st condition of equilibrium:

∑F_x = 0

∑F_y = 0

Now,

F - 40 - 30 - 50 = 0

F - 120 = 0

F = 120 N

(ii) Now by applying 2^nd condition of Equilibrium

∑𝜏 = 0

F x AD - W x AC - 50 x AB = 0

120 x X -30 x 5 - 50 x 10 = 0

120X - 650 = 0

120X = 650

X = ⁶⁵⁰/₁₂₀

X = 5.41 m

Ans: Therefore, the position of the fulcrum is 5.41 cm.

12) A force of 25 N acts on a body. If moment of arm is 2 m, find the value of torque?
Ans: Solution:
Data:

• Force = F = 25 N
• Moment of arm = d = 2 m

Required Data:

• Torque = 𝜏 = ?

Formula:

𝜏 = F x d

Calculation:

𝜏 = 25 x 2

𝜏 = 50 Nm

Ans: Therefore, the value of torque is 50 Nm.

13) A force is applied perpendicularly on a door 4 meters wide which requires a torque of 120 Nm to open it. What will be minimum force required?
Ans: Solution:
Data:

• Moment of arm = d = 4 m
• 𝜏 = 120 Nm

Required Data:

• Force = F = ?

Formula:

𝜏 = F x d

Calculation:

120 = F x 4

120 = 4F

F =  ¹²⁰/₄  = 30 N

Ans: Therefore, minimum force required is 30 N.

14) What is the moment of the couple of 10 N acting at the extremities of a rod 5 m long as shown in figure. How can this couple be balanced?
Ans: Solution:
Data:

• Force = F = 10 N Couple arm
• = d = 5 m

Required Data:

• Moment of couple or torque = 𝜏 = ?

Formula:

𝜏 = F x d

Calculation:

𝜏 = 10 x 5 = 50 Nm

Ans: Therefore, the moment of the couple or torque is 50 Nm.