Q1. A 100 kg crate is resting on a horizontal surface. If the coefficient of static friction is 0.5, what is the maximum static friction force?
Solution:
The maximum static friction force Ff = μs * N = 0.5 * (100 kg * 9.81 m/s²) = 490.5 N.
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Q2. A block on an inclined plane has a mass of 10 kg and the angle of inclination is 30 degrees. What is the force of friction if the coefficient of kinetic friction is 0.2?
Solution:
The normal force N = mg * cos(θ) = 10 kg * 9.81 m/s² * cos(30°) ≈ 84.87 N. The force of friction Ff = μk * N = 0.2 * 84.87 N ≈ 16.97 N.
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Q3. A box weighing 300 N is pushed with a force of 150 N. If the coefficient of kinetic friction is 0.4, will the box move?
Solution:
The force of kinetic friction Ff = μk * N = 0.4 * 300 N = 120 N. Since the applied force (150 N) is greater than the frictional force (120 N), the box will move.
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Q4. If a car is moving at a speed of 20 m/s and the coefficient of friction between the tires and the road is 0.7, what is the maximum deceleration the car can achieve?
Solution:
The maximum deceleration a car can achieve is given by a = μg, where g = 9.81 m/s². Thus, a = 0.7 * 9.81 m/s² ≈ 6.87 m/s².
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Q5. A 250 kg object is subjected to a horizontal force of 600 N. If the coefficient of friction is 0.5, will the object move?
Solution:
The maximum static friction force Ff = μs * N = 0.5 * (250 kg * 9.81 m/s²) = 1226.25 N. Since the applied force (600 N) is less than the frictional force (1226.25 N), the object will not move.
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Q6. If a 150 kg object is on a slope of 15 degrees, what is the normal force acting on it?
Solution:
The normal force N = mg * cos(θ) = 150 kg * 9.81 m/s² * cos(15°) ≈ 1430 N.
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Q7. A 200 kg object is on a flat surface with a coefficient of kinetic friction of 0.3. What is the force of friction acting on the object?
Solution:
The force of friction Ff = μk * N = 0.3 * (200 kg * 9.81 m/s²) = 588.6 N.
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Q8. What is the coefficient of static friction if a 50 kg box requires a force of 200 N to start moving?
Solution:
The coefficient of static friction (μs) can be calculated using the formula F = μs * N, where N is the normal force. Here, N = mg = 50 kg * 9.81 m/s² = 490.5 N. Thus, μs = F/N = 200 N / 490.5 N ≈ 0.4.
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Q9. What is the angle of inclination at which the force of static friction equals the component of weight acting down the slope for a coefficient of static friction of 0.4?
Solution:
The angle θ can be found using tan(θ) = μs. Thus, θ = arctan(0.4) ≈ 21.8 degrees.
