Turning Effect of Forces Physics Class 9 short question, Mcqs, and Numerical Problems.
Mcqs
Table of Contents
i) Two equal but unlike parallel forces having different lines of action produce
- A. a torque
- B. a couple
- C. equilibrium
- D. neutral equilibrium
ii) The number of forces that can be added by head to tail rule is:
- A. 2
- B. 3
- C. 4
- D. any number
iii) The number of perpendicular components of a force are:
- A. 1
- B. 2
- C. 3
- D. 4
iv) A force of 10 N is making an angle of 30° with the horizontal. Its horizontal component will be :
- A. 4 N
- B. 5 N
- C. 7 N
- D. 8.7 N
v) A couple is formed by
- A. two forces perpendicular to each other
- B. two like parallel forces
- C. two equal and opposite forces in the same line
- D. two equal and opposite forces not in the same line
vi) A body is in equilibrium when it:
- A. acceleration is uniform
- B. speed is uniform
- C. speed and acceleration are uniforms
- D. acceleration is zero
vii) A body is in neutral equilibrium when its center of gravity:
- A. is at its highest position
- B. is at the lowest position
- C. keeps its height if displaced
- D. is situated at its bottom
viii) Racing cars are made stable by
- A. increasing their speed
- B. decreasing their mass
- C. lowering their center of gravity
- D. decreasing their width
Questions
Q.2) Define the following:
i) resultant vector
ii) torque
iii) center of mass
iv) center of gravity
Answer:
i) RESULTANT VECTOR
The resultant vector is the vector that is obtained when two or more vectors are added up.
ii) TORQUE
The turning effect of a force is called torque or moment of force.
Mathematically,
Where r is the distance between the axis and point of action of force and F is the applied force.
Examples: turning a pencil in a sharpener, turning the stopcock of a water tap, using a spanner, and opening and closing a door are some examples where a force produces a turning effect.
Torque is a vector quantity and its direction can be found by using the right-hand rule. The S.I unit of torque is Nm.
iii) CENTER OF MASS
Centre of mass of a system is such a point where an applied force causes motion without rotation.
iv) CENTER OF GRAVITY
A point where the whole weight of the body appears to act vertically downward is called the center of gravity of the body.
Q.3) Differentiate the following:
i) like and unlike forces
ii) torque and couple
iii) stable and neutral equilibrium
Answer:
i) LIKE AND UNLIKE FORCES
Forces that are parallel and acting in the same direction are called parallel forces.
Forces that are parallel and acting in opposite directions are called, unlike parallel forces.
ii) TORQUE AND COUPLE
The turning effect produced by a force on a body is called torque. It is a product of applied force and the perpendicular distance between the axis of rotation and line of action of force.
The formula of torque is;
Two parallel forces equal in magnitude but opposite in direction trying to rotate a body in the same direction constitute a couple.
iii) STABLE AND NEUTRAL EQUILIBRIUM
A body is said to be in stable equilibrium if the body is slightly displaced and returns to its original position.
If a body maintains its new position when it is disturbed from its previous position, the body is said to be in neutral equilibrium.
Q.4) How head to tail rule helps find the resultant of forces?
Answer:
Head to tail rule helps in the addition of vectors. Hence used to find the resultant of forces. To add the vectors by head to tail rule, join the head of the first vector to the tail of the second vector. Similarly, draw the next vector such that its tail coincides with the head of the second vector and so on. Now draw a vector such that its tail is at the tail of the first vector and its head on the head of the last vector. This final vector is known as the resultant vector and this method of addition of vectors is known as the head to tail rule.
Mathematically, the resultant of the forces FA, FB, FC … is given by;
R = FA + FB +FC…
For two vectors
R = A + B
In figure, two vectors added by head to tail rule.
Q.5) How can a force be resolved into its perpendicular components?
Answer:
Splitting up a force into two mutually perpendicular components is called the resolution of that force and such mutually perpendicular components are known as rectangular components.
Consider a force F represented by line OA making an angle ϴ with an x-axis. Draw a perpendicular AB on the x-axis from A. According to head to tail rule, OA is the resultant vector of OB and BA.
Thus
OA = OB + BA
From the figure
F = Fx + Fy
Fx and Fy can be calculated by
Q.6) When a body is said to be in equilibrium?
Answer:
Q.7) Explain the first condition of equilibrium.
Answer:
FIRST CONDITION OF EQUILIBRIUM
“The sum of all the forces acting on a body is zero.”
Let n number of forces F 1, F2, F3,…, Fn which are acting on a body. If the body satisfies the first condition of equilibrium,
In terms of x and y components, the first condition can be written as
Examples: A picture frame hanging on a wall and a paratrooper coming down with constant velocity, both satisfy the first condition of equilibrium.
Q.8) Why is there a need for a second condition for equilibrium if the body satisfies the first condition for equilibrium?
Answer:
The first condition of equilibrium does not make sure that a body is in equilibrium.
For example, consider a body as shown in the figure in which it is pulled by two forces. These two forces are equal in magnitude but opposite in direction to each other and act along the different lines, so the body rotates due to a couple. In this situation, the body is not in equilibrium although the first condition of equilibrium is satisfied. It is because the body rotates. So, another condition for equilibrium needs to be satisfied and this condition is known as the second condition of equilibrium. According to this, the sum of all the torques acting on the body is zero.
Mathematically;
Q.9) What is the second condition for equilibrium?
Answer:
SECOND CONDITION OF EQUILIBRIUM
The sum of all the torque acting on the body is zero.
Mathematically;
For example, a child’s seesaw. The torque produced due to the weight of the child at the right side is clockwise and the torque produced due to weight of the child at the left side is anticlockwise. So these two torques are opposite in direction to each other. If both the torques are equal in magnitude, then the sum of these two torques is zero. Thus, the second condition of equilibrium satisfied.
Q.10) Give an example of a moving body that is in equilibrium.
Answer:
A car moving with uniform velocity is an example of a moving body in equilibrium. The force of the engine acts in a forwarding direction while the force of friction between the road and the tires acts backward. These two forces, being equal and opposite, cancel the effect of each other and the car moves with uniform velocity.
Q.11) Think of a body that is at rest but not in equilibrium.
Answer:
It’s impossible that a body is at rest but not in equilibrium. Because, whenever a body is at rest it means that, the sum of all the forces acting on the body is zero. So, the first condition of equilibrium is always satisfied. Therefore, all bodies which are at rest are in equilibrium.
Q.12) Why a body cannot be in equilibrium due to a single force acting on it?
As we know that, the first condition of equilibrium is;
“The net force acting on the object must be zero.”
If there is a single force acting on the body, the net force cannot be zero as there is no other force to balance this single force. Thus the first condition of equilibrium will not be satisfied. Hence, a body cannot be in equilibrium due to a single force acting on it.
Q.13) Why is the height of vehicles kept as low as possible ?
Answer:
As a whole, the weight of a body acts on the center of gravity and the stability of objects is related to the center of gravity point. If the center of gravity point is near to earth, then the object is more stable. Due to this reason, the height of vehicles kept low as much as possible to lower the center of gravity point, in order to increase their stability.
Q.14) Explain what is meant by stable, unstable and neutral equilibrium. Give one example in each case.
Answer:
STABLE EQUILIBRIUM
A body is said to be in stable equilibrium, if the body is slightly displaced and it tends to return to its original position.
Example: A ball located at the bottom of a spherical deepening, is in a state of stable equilibrium. When we place it in a high position, then after some time it comes to the bottom again.
UNSTABLE EQUILIBRIUM
If a body does not return to its previous position when it sets free after a small disturbance, it is said to be in unstable equilibrium.
Example: A ball located at the top of a spherical projection is an example of unstable equilibrium.
NEUTRAL EQUILIBRIUM
If a body maintains its position when it is disturbed from the previous position,it is said to be in neutral equilibrium.
Example: The ball shown below will come to rest again if moved to another location on the surface shown.
Numerical Problems
4.1 Find the resultant of the following forces:
- 10 N along x-axis
- 6 N along y-axis and
- 4 N along negative x-axis. (8.5 N making 45⁰ with x-axis)
Solution: Fx = Net force along x-axis = 10.4 = 6 N
Fy = Force along y-axis = 5 N
Magnitude of the resultant force = F = ?
F = √Fx2 + Fy2
F = √(6)2 + (6)2
F = √36 + 36
= √72 = 8.5 N
Now, θ = tan-1 = Fy / Fx
θ = tan-1 = 6 / 6
θ = tan-1 (1)
θ = 45⁰ with x-axis
4.2 Find the perpendicular components of a force of 50 N making an angle of 30⁰ with x-axis. (43.3 N, 25 N)
Solution: Force F = 50 N
Angle θ = 30⁰
Fx = ? and Fy = ?
Fx = F cos θ
Fx = 50 × cos 30
= 50 N × 0.866 (˙.˙ cos 30⁰ = 0.866)
Fx = 43.3 N
Similarly, Fy = F sin θ
Fy = 50 × 0.5 (˙.˙ sin 30⁰ = 0.5)
Fy = 25 N
4.3 Find the magnitude and direction of a force, if its x-component is 12 N and y-component is 5 N. (13 N making 22.6⁰ with x-axis)
Solution: Fx = 12 N
Fy = 5 N
- Magnitude of the force = F = ?
- Direction of the force = θ = ?
F = √Fx2 + Fy2
F = √(12)2 + (5)2
F = √144+25
F = 13 N
(ii) θ = tan-1 = Fy / Fx
θ = tan-1 = 12 / 5
θ = tan-1 (2.4)
θ = 22.6⁰ with x-axis
4.4 A force of 100 N is applied perpendicularly on a spanner at a distance of 10 cm from a nut. Find the torque produced by the force. (10 Nm)
Solution: Force = F = 100 N
Distance = L = 10 cm = 0.1 m
Torque Τ = ?
Torque Τ = F × L
= 100 N × 0.1 m
= 10 Nm
4.5 A force is acting on a body making an angle of 30⁰ with the horizontal. The horizontal component of the force is 20 N. Find the force. (23.1 N)
Solution: Angle θ = 30⁰ (with x-axis)
Horizontal component of force Fx = 20 N
Force F = ?
Fx = F cos θ
20 N = F cos 30⁰
20 N = F × 0.866 (˙.˙ cos 30⁰ = 0.866)
F = 20 N / 0.866 = 23.09
F = 23.1 N
4.6 The steering of a car has a radius 16 cm. Find the torque produced by a couple of 50 N. (16 Nm)
Solution: Radius = r = L = 16 cm = 16/100 m = 0.16 m
Couple arm = L = 16 cm = 16/100 m = 0.16 m
Force = F = 50 N
Torque Τ = ?
Torque Τ = F × L
= 50 N × (2 × 0.16)
= 16 Nm
4.7 A picture frame is hanging by two vertical strings. The tensions in the strings are 3.8 N and 4.4 N. Find the weight of the picture frame. (8.2 N)
Solution: Tension T1 = 3.8 N
Tension T2 = 4.4 N
Weight of the picture frame = w = ?
When the picture is in equilibrium, then
∑ Fx = 0 and ∑ Fy = 0
Therefore T – w = 0
Or (T1 + T2) – w = 0
T1 + T2 = w
3.8 + 4.4 = w
W = 8.2 N
4.8 Two blocks of mass 5 kg and 3 kg are suspended by the two strings as shown. Find the tension in each string. (80 N, 30 N)
Solution: Mass of large block = M = 5 kg
Mass of large block = m = 3 kg
Tension produced in each string = T1 = ? and T2 = ?
T1 = w1 + w2
T1 = Mg + mg
T1 = (M + m)g
T1 = (3+5) × 10
= 8 × 10
= 80 N
Also, T2 = mg
T2 = 3 × 10 = 30 N
4.9 A nut has been tightened by a force of 200 N using 10 cm long spanner. What length of a spanner is required to loosen the same nut with 150 N force? (13.3 cm)
Solution: Force = F1 = 200 N
Length = L1 = 10 cm = 10 / 100 = 0.1 m
Length of the spanner to tighten the same nut:
Force = F2 = 150 N
Length = L2 = ?
Since Τ1 = Τ2
F1 × L1 = F2 × L2
200 × 0.1 = 150 × L2
20 = 150 × L2
L2 = 20 / 150 = 0.133 m = 0.133 × 100 = 13.3 cm
4.10 A block of mass 10 kg is suspended at a distance of 20 cm from the center of a uniform bar 1 m long. What force is required to balance it at its center of gravity by applying the force at the other end of the bar? (40 N)
Solution: Mass of the block = m = 10 kg
Length of the bar = l = 1 m
Moment arm of w1 = L1 = 20 cm = 0.2 m
Moment arm of w2 = L2 = 50 cm = 0.5 m
Force required to balance the bar F2 = ?
By applying principle of moments:
Clockwise moments = anticlockwise moments
F1 × L1 = F2 × L2
mg × L1 = F2 × L2
(10 ×10 ) × 0.2 = F2 × 0.5
20 = F2 × 0.5
F2 = 20 / 0.5 = 200 / 5 = 40 N
