A particle of mass 9kg rests on a rough plane inclined at 30degrees to the horizontal. The coefficient of friction between the particle and the plane is 1/3. A horizontal force is applied to the particle so that the particle just begins to move upwards. Find the magnitude of this horizontal force.
To find the magnitude of the horizontal force required for the particle to just begin moving upwards, we need to consider the forces acting on the particle.
1. Weight: The weight of the particle is given by the formula W = mg, where m is the mass of the particle (9 kg) and g is the acceleration due to gravity (9.8 m/s²). Therefore, the weight W = 9 kg × 9.8 m/s² = 88.2 N.
2. Normal force: The component of the weight perpendicular to the inclined plane is balanced by the normal force. The normal force N = mg cosθ, where θ is the angle of inclination (30 degrees). Therefore, the normal force N = 9 kg × 9.8 m/s² × cos(30°) ≈ 75.3 N.
3. Friction force: The friction force acting on the particle is given by the formula f = μN, where μ is the coefficient of friction (1/3) and N is the normal force (75.3 N). Therefore, the friction force f = (1/3) × 75.3 N ≈ 25.1 N.
For the particle to just begin moving upwards, the applied horizontal force should overcome the force of friction. Therefore, the magnitude of the required horizontal force is equal to the friction force, which is approximately 25.1 N.
To solve this problem, we need to consider the forces acting on the particle and use Newton's second law of motion.
First, let's draw a free-body diagram of the particle on the inclined plane:
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______|______ <- inclined plane
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|------|-------| <- particle
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friction
Now let's analyze the forces acting on the particle:
1. Weight (W): This force acts vertically downwards and can be calculated as W = mg, where m is the mass of the particle and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. Normal force (N): This force acts perpendicular to the plane and counterbalances the component of weight acting parallel to the plane. The normal force can be calculated as N = mg * cosθ, where θ is the angle of inclination.
3. Friction force (f): This force opposes the motion of the particle along the inclined plane. The friction force can be calculated as f = μN, where μ is the coefficient of friction between the particle and the plane.
4. Applied force (F): This horizontal force is what we want to find. It acts parallel to the inclined plane and causes the particle to move upwards.
Now, the condition for the particle to just begin moving upwards is that the force due to friction is equal to the applied force (F). Mathematically, we can represent this as:
f = F
Substituting the equations for f and N, we have:
μN = F
μ(mg * cosθ) = F
Given that the coefficient of friction (μ) is 1/3, the mass of the particle (m) is 9 kg, and the angle of inclination (θ) is 30 degrees, we can substitute these values into the equation:
(1/3)(9 * 9.8 * cos30) = F
Calculating this expression, we find:
(1/3)(9 * 9.8 * 0.866) ≈ 44.212
Therefore, the magnitude of the horizontal force (F) required to just begin moving the particle upwards is approximately 44.212 Newtons.