🔹 Topic: Static Friction and Limiting Friction | MCQs: 50 🔹
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1. A block weighs W is held against a vertical wall by applying a horizontal force F. The minimum value of F needed to hold the block is
Correct Answer: (c) Greater than W
To hold the block against the vertical wall, the frictional force (μF) must balance the weight (W). For minimum F, we have μF = W ⇒ F = W/μ. Since μ < 1, F must be greater than W.
2. Maximum value of static friction is called
Correct Answer: (a) Limiting friction
The maximum value of static friction that comes into play when a body just starts moving over another surface is called limiting friction.
3. Pulling force making an angle θ to the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is λ, then the magnitude of force required to move the body is equal to
Correct Answer: (a) \( \frac{W \sin \lambda}{\cos(\theta - \lambda)} \)
The minimum force required to move the block is given by \( F = \frac{W \sin \lambda}{\cos(\theta - \lambda)} \), where λ is the angle of friction and θ is the angle of applied force.
4. A uniform rope of length l lies on a table. If the coefficient of friction is μ, then the maximum length l₁ of the part of this rope which can overhang from the edge of the table without sliding down is
Correct Answer: (a) \( \frac{\mu l}{1 + \mu} \)
For equilibrium, the weight of the hanging part must be balanced by the frictional force on the part lying on the table. If l₁ is the hanging length, then \( l_1 \rho g = \mu (l - l_1) \rho g \), solving gives \( l_1 = \frac{\mu l}{1 + \mu} \).
5. Which of the following statements is not true
Correct Answer: (c) Rolling friction is greater than sliding friction
Rolling friction is actually much smaller than sliding friction. This is why we use wheels and ball bearings to reduce friction in machines.
6. A block of 1 kg is stopped against a wall by applying a force F perpendicular to the wall. If μ = 0.2 then minimum value of F will be
Correct Answer: (b) 49 N
The frictional force μF must balance the weight of the block (mg = 1 × 9.8 = 9.8 N). So, 0.2 × F = 9.8 ⇒ F = 9.8/0.2 = 49 N.
7. A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table surface is 0.25, then the maximum fraction of length of the chain, that can hang over one edge of the table is
Correct Answer: (a) 20%
Using the formula \( \frac{l_1}{l} = \frac{\mu}{1 + \mu} = \frac{0.25}{1.25} = 0.2 \) or 20%.
8. Work done by a frictional force is
Correct Answer: (d) All of the above
Work done by friction can be:
- Negative when it opposes motion (most common case)
- Positive when it helps motion (e.g., feet pushing backward to move forward)
- Zero when there's no relative motion (static friction)
9. A uniform chain of length L changes partly from a table which is kept in equilibrium by friction. The maximum length that can withstand without slipping is l, then coefficient of friction between the table and the chain is
Correct Answer: (a) \( \frac{l}{L - l} \)
For equilibrium, weight of hanging part (l) must be balanced by friction on the part on table (L-l): \( l \rho g = \mu (L - l) \rho g \) ⇒ \( \mu = \frac{l}{L - l} \).
10. When two surfaces are coated with a lubricant, then they
Correct Answer: (b) Slide upon each other
Lubricants reduce friction between surfaces, making them slide more easily upon each other.
11. A 20 kg block is initially at rest on a rough horizontal surface. A horizontal force of 75 N is required to set the block in motion. After it is in motion, a horizontal force of 60 N is required to keep the block moving with constant speed. The coefficient of static friction is
Correct Answer: (a) 0.38
Coefficient of static friction μₛ = Fₛ/N = Fₛ/mg = 75/(20×9.8) ≈ 0.38.
12. To avoid slipping while walking on ice, one should take smaller steps because of the
Correct Answer: (c) Friction of ice is small
Ice has very low friction. Smaller steps reduce the horizontal component of force, making it less likely to exceed the small frictional force available.
13. A box is lying on an inclined plane what is the coefficient of static friction if the box starts sliding when an angle of inclination is 60°
Correct Answer: (b) 1.732
At the angle of repose, μ = tanθ = tan60° = √3 ≈ 1.732.
14. Which one of the following is not used to reduce friction
Correct Answer: (c) Sand
Sand actually increases friction (used on icy roads). Oil, ball bearings, and graphite are all used to reduce friction.
15. If a ladder weighing 250 N is placed against a smooth vertical wall having coefficient of friction between it and floor is 0.3, then what is the maximum force of friction available at the point of contact between the ladder and the floor
Correct Answer: (a) 75 N
Maximum frictional force = μN = 0.3 × 250 = 75 N (normal reaction equals weight for vertical equilibrium).
16. A body of mass 2 kg is kept by pressing to a vertical wall by a force of 100 N. The coefficient of friction between wall and body is 0.3. Then the frictional force is equal to
Correct Answer: (b) 20 N
Frictional force balances the weight of the body: f = mg = 2 × 9.8 ≈ 19.6 N ≈ 20 N. The maximum possible friction (μN = 0.3×100 = 30 N) is not reached here.
17. A uniform metal chain is placed on a rough table such that one end of chain hangs down over the edge of the table. When one-third of its length hangs over the edge, the chain starts sliding. Then, the coefficient of static friction is
Correct Answer: (a) \( \frac{1}{2} \)
At limiting equilibrium: weight of hanging part (L/3) = friction on part on table (2L/3). So \( \frac{L}{3} \rho g = \mu \frac{2L}{3} \rho g \) ⇒ μ = 1/2.
18. A lift is moving downwards with an acceleration equal to acceleration due to gravity. A body of mass m kept on the floor of the lift is pulled horizontally. If the coefficient of friction is μ, then the frictional resistance offered by the body is
Correct Answer: (c) Zero
In a freely falling lift (a = g), the body experiences weightlessness (apparent weight = 0). Friction = μN = μ × 0 = 0.
19. The angle of friction is the angle between
Correct Answer: (c) The resultant of normal reaction and frictional force and the normal reaction
The angle of friction (λ) is defined as the angle between the resultant reaction (R) and the normal reaction (N), where tanλ = μ (coefficient of friction).
20. The force required to just move a body up the inclined plane is double the force required to just prevent it from sliding down. If θ is angle of inclination and μ is coefficient of friction, then
Correct Answer: (b) tanθ = 3μ
Force to move up: \( F_1 = mg(\sinθ + μ\cosθ) \)
Force to prevent sliding down: \( F_2 = mg(\sinθ - μ\cosθ) \)
Given \( F_1 = 2F_2 \) ⇒ \( \sinθ + μ\cosθ = 2(\sinθ - μ\cosθ) \) ⇒ \( \sinθ = 3μ\cosθ \) ⇒ \( \tanθ = 3μ \).
Force to prevent sliding down: \( F_2 = mg(\sinθ - μ\cosθ) \)
Given \( F_1 = 2F_2 \) ⇒ \( \sinθ + μ\cosθ = 2(\sinθ - μ\cosθ) \) ⇒ \( \sinθ = 3μ\cosθ \) ⇒ \( \tanθ = 3μ \).
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