Moment of Forces MCQ Quiz - Objective Question with Answer for Moment of Forces - Download Free PDF

Explanation:

1. Mass is a fundamental property of matter that does not change due to the application of force.
2. The mass of an object is a measure of the amount of matter it contains and is unaffected by the forces acting on it.
3. For example, even if a car accelerates or changes direction, its mass remains constant. The only time mass would change is if there is a physical transformation of the substance (like burning fuel, which reduces the mass of the fuel).

 Additional Information

A force is any interaction that, when unopposed, will change the motion of an object. It can cause objects to start moving, stop moving, or change direction.

Forces have both magnitude and direction, making them vector quantities. The effects of force on a body depend on:

1. The type of force: Whether it is gravitational, normal, frictional, tension, or applied force.

2. The magnitude of the force: A greater force generally causes a larger effect.

3. The point of application: The place where the force is applied can affect how the object moves (e.g., pushing from the center or edge of an object).

4. The mass of the object: According to Newton's second law, a more massive object requires a greater force to produce the same acceleration.

Explanation:

The moment of a force (also called torque) is a measure of the tendency of the force to rotate an object about a specific point or axis.

It is mathematically defined as:

Moment (M)=Force (F)×Perpendicular Distance (d)

Where:

• Force (F) is the magnitude of the applied force.

• Perpendicular Distance (d) is the shortest distance from the point of rotation to the line of action of the force.

 Additional InformationTypes of Moments:

1. Clockwise Moment:

   • Causes rotation in the clockwise direction.

2. Counter-clockwise Moment:

   • Causes rotation in the anti-clockwise direction.

Principle of Moments (Varignon’s Theorem):

• States that if a number of forces act on a body, the total moment about any point is equal to the sum of the moments of the individual forces.

Concept:

Since the system is in equilibrium under three concurrent forces, we apply Lami’s Theorem:

\( \frac{T_{AC}}{\sin(\angle BCF)} = \frac{T_{BC}}{\sin(\angle ACF)} = \frac{W}{\sin(\angle ACB)} \)

Given:

Weight (W) = 15 N

Angle between rope AC and vertical = 30°

Angle between rope BC and vertical = 45°

Angle between ropes AC and BC = 30° + 45° = 75°

sin(30°) = 0.5, sin(75°) = cos(15°) = 0.966

Calculation:

Using Lami’s Theorem:

\( \frac{T_{BC}}{\sin(30°)} = \frac{15}{\sin(75°)} \)

Substitute values:

\( \frac{T_{BC}}{0.5} = \frac{15}{0.966} \)

Cross-multiplying:

\( T_{BC} = \frac{15 \times 0.5}{0.966} = 7.76 \, N \)

Concept:

Moment of Force:- The tendency of forces is not only to move a body but also to rotate the body.

This rotational tendency of a force is called a moment of force.

It can be defined as "The product of force and perpendicular distance from the point to the line of action of the force is called the moment of a force about that point.

The SI unit of the moment of force is Newton-meter (N-m) while in CGS it can be written as newton centimeter (N-cm)

Formula Used:

M = F ×  d

Where, M = Moment

F = Force

d = perpendicular distance

Explanation:

Geometric Representation of Moment of Force:-

The moment of a force about a point is equal to twice the area of the triangle so formed by talking the point as the vertex of the triangle and the line as the base of the triangle.

Let O be the point about which moment of the force is to be calculated.

Then, Moment of the force about O,

⇒ M = F × OC

Moment of the force about O,

⇒ M = AB × OC  [F = AB]

Now multiplied and divide by 2,

⇒ M = 2 × \(\frac{1}{2} \) × AB × OC

Area of triangle = 1/2 × base × height

⇒ M = 2 × △OAB

∴ The value of λ is 2

Concept:

The resultant of two or more than two vectors are always vector.

Explanation:

Let us consider a Hexagon ABCDEF with centre O.
Let us consider AB and AE as axes.If K is the proportionality constant .
Then the forces along the sides AB,BC,CD,DE,,EF and FA are of magnitudes K,2K,3K,4K, 5K and 6K respectively.
Let R be the resultant vector making an angle θ with AX and at a distance d from the center of hexagon.
Resolving the forces along Ax
we get,
R Cosθ=K+2K Cos 60° +3K cos120° -4K-5K Cos 60° -6K cos 120°
= K + 2k.1/2 + 3k(-1/2) - 4k - 5k.1/2 - 6K(-1/2)
= K+K-3K/2-4K-5K/2+3K
= -3KCOSθ
= -3K -----------(1)
Resolving along AY ,
we get
R sinθ = 2K Sin60° +3K sin120°- 5Ksin60° - 6Ksin120°
= 2Ksin60°+3K sin60° -5ksin60°-6ksin60°
= - 6Ksin60°
= - 6K.√3/2
= - 3√3 .K---------------(2)
R sinθ=-3√3.K
Squaring equation 1 and 2
we get :
R²=9K² +27K²
=36K²
R=6K
Here K = 1

So correct option (2) 6N is the correct answer.

Concept:

\(Moment = P × OC\)

And

\(Area\;of\;triangle = \frac{1}{2} × AB × OC\)

                            \( = \frac{1}{2} × P × OC\)

                            = \(\frac{1}{2}\)× moment

∴ Moment = Twice the area of a triangle

Concept:

Conditions for the system to be in equlibrium

ΣF_x = 0, ΣF_y = 0, ΣM = 0

Calculation:

Given:

m = 2 kg, Assume g = 10 m / s²

Lager will be the moment, smaller will be the force required to lift the rod. Hence, Applying Moment about 0 cm point we get.

w × 50 = F × 100

m × g × 50 = F × 100

2 × 10 × 50 = F × 100

F = 10 N

Explanation:

Type of Connection	Reaction	Number of Unknowns
Weight legs link		One - The reaction is a force that acts in the direction of the link
Rollers		One - The reaction is a force that act perpendicular to the surface at point of contact.
Pin or Hinge		Two - The reaction are two force components
Guided rollar/ Fixed connected collar		Two - The reactions are a force and a moment
Fixed support		Three - The reactions are two forces and a moment
Pin connected collar		One - The reaction is a force that acts perpendicular to the surface at the point of contact

Explanation:

To determine the couple force exerted by the blade edges on the screw slot, we need to calculate the force applied by the screwdriver blade and then multiply it by the lever arm.

The formula for torque is given by:

Torque = Force x Lever Arm

In this case, the torque is 4 Nm and the width of the screwdriver blade is 5 mm. However, we need to convert the width of the blade to meters before proceeding with the calculation. 1 mm is equal to 0.001 meters.

Width of the screwdriver blade = 5 mm = 5 x 0.001 m = 0.005 m

Now we can rearrange the formula for torque to solve for force:

Force = Torque / Lever Arm

Force = 4 Nm / 0.005 m = 800 N

Therefore, the couple force exerted by the blade edges on the screw slot is 800 N.

So, the correct answer is option 2.

Explanation:

Equilibrium of planar forces:

If the forces acting on the free-body are non-concurrent then the equivalent resultant will be a single force acting at a common point and a moment about the same point. The effect of such a force system will be to translate the body as well as to rotate it. Hence, for equilibrium to exist, both the force and moment must be null vectors.

i.e ΣF = 0, ΣM = 0

When the forces acting on a body lie in a plane (X - Y plane) but are non-concurrent, the body will have rotational motion perpendicular to the plane in addition to the translational motion along the plane, Hence necessary and sufficient condition for static equilibrium are,

ΣF_x = 0, ΣF_y = 0, ΣM_z = 0

Concept:

Parallelogram Law of Forces:

If two forces, acting at a point be represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point. 

Resultant force Fr, of any two forces F1 and F2 with an angle θ between them, can be given by vector addition as

\(F_r^2 = F_1^2 + F_2^2 + 2{F_1}{F_2}\cos θ\)

\(F_r = \sqrt {F_1^2 + F_2^2 + 2{F_1}{F_2}\cos θ}\)'

Calculation:

Given:

Two forces P and Q are acting at a particle. The angle between the forces is θ and the resultant of the forces is R:

If the resolved part of R in the direction of P is 2P:

Resolved part of R = Q cosθ + P

i.e. 2P = Q cosθ + P

P =  Q cosθ

Explanation:

When we draw a free body diagram we remove all the supports and force applied due to that support are drawn and force or moment will apply in that manner so that it resists forces in any direction as well as any tendency of rotation.

Following is the conclusion:

Option (B) is wrong because the ground support is shown which we never show in FBD.

Option (C) is wrong because the X component is not shown.

Option (D) is wrong because the moment produced due to the eccentric force is not shown.

Concept:

The moment about the resultant is zero.

Calculation:

Let the two forces be F₁ and F₂

Given:

F₁ + F₂ = 60 N        (1) (They are like forces and parallel)

Taking the moment about the resultant

\(\begin{array}{l} \mathop F\nolimits_1 \times 10 - \mathop F\nolimits_2 \times 20 = 0\\ \mathop F\nolimits_1 = 2\mathop F\nolimits_2 \end{array}\)

By solving, we get

F₂ = 20 N

F₁ = 40 N

Explanation:

Varignon’s Principle of moments (or the law of moments):

It states, “If a number of coplanar forces are acting simultaneously on a particle, the algebraic sum of the moments of all the forces about any point is equal to the moment of their resultant force about the same point.”

i.e. R × r = F1 × r1 + F2 × r2

Therefore from the above principle, we can conclude that the dynamic effect of the resultant acting on a rigid body in equilibrium will be the same effect when acted by a given system of forces.

Explanation:

Varignon’s Principle of moments (or the law of moments)

• It states, “If a number of coplanar forces are acting simultaneously on a particle, the algebraic sum of the moments of all the forces about any point is equal to the moment of their resultant force about the same point.”

MO' = R × r = F1 × r1 + F2 × r2

Additional Information

Lami's theorem

• Lami's theorem states that if three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces. Consider three forces FA, FB, FC acting on a particle or rigid body making angles α, β, and γ with each other.

Therefore, \(\frac{{{F_A}}}{{sin\alpha }} = \frac{{{F_B}}}{{sin\beta }} = \frac{{{F_C}}}{{sin\gamma }}\)

D’ Alembert's principle:

• It states that a moving body can be brought to equilibrium by adding inertia force to the system whose magnitude is equal to the product of mass and acceleration.
• This inertia force is in the opposite direction to that of acceleration, i.e. F = ma

Polygon Law of Forces

• It is an extension of Triangle Law of Forces for more than two forces, which states, “If a number of forces acting simultaneously on a particle, be represented in magnitude and direction, by the sides of a polygon taken in order; then the resultant of all these forces may be represented, in magnitude and direction, by the closing side of the polygon, taken in the opposite order.”