A particle of mass 10 kg is placed on an inclined plane at an angle of 30 degrees. The coefficient of static friction is 0.4. Determine the forces involved and whether the particle remains at rest.

Question

Asked on May 11, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-05-11T14:19:42.647Z

Welcome back aborkugyaanku4@gmail.com — been a while! Let's pick up where you left off. To solve this problem, we first need to determine the components of the gravitational force acting on the particle parallel and perpendicular to the inclined plane, and then calculate the maximum static friction force. Given: Mass m = 10 kg Angle of inclination = 30^ Coefficient of static friction = 0.4 Acceleration due to gravity g = 10 m/s^2 Step 1: Calculate the components of the gravitational force and the normal force. The weight of the particle is W = mg. W = 10 kg × 10 m/s^2 = 100 N The component of the weight parallel to the plane (acting down the plane) is W_. W_ = W = 100 N × 30^ = 100 N × 0.5 = 50 N The component of the weight perpendicular to the plane is W_. W_ = W = 100 N × 30^ = 100 N × sqrt(3)2 ≈ 100 N × 0.866 = 86.6 N The normal force N is equal to the perpendicular component of the weight. N = W_ = 86.6 N The maximum static friction force f_s is given by N. f_s = N = 0.4 × 86.6 N = 34.64 N i) To hold the particle at rest: The particle tends to slide down the plane due to W_ = 50 N. Since W_ (50 N) is greater than the maximum static friction force f_s (34.64 N), an external force F_1 must be applied up the plane to prevent it from sliding down. In this case, the friction force acts up the plane, opposing the downward motion. Step 2: Calculate the force F_1 required to hold the particle at rest. For equilibrium, the sum of forces parallel to the plane must be zero. F_1 + f_s = W_ F_1 = W_ - f_s F_1 = 50 N - 34.64 N F_1 = 15.36 N The force required to hold the particle at rest is 15.36 N. ii) To just make the particle move up the plane: An external force F_2 must be applied up the plane. In this case, the particle is about to move up, so the friction force f_s acts down the plane, opposing the upward motion. The component of gravity W_ also acts down the plane. Step 3: Calculate the force F_2 required to just make the particle move up the plane. For impending upward motion, the sum of forces parallel to the plane must be zero. F_2 = W_ + f_s F_2 = 50 N + 34.64 N F_2 = 84.64 N The force required to just make the particle move up the plane is 84.64 N. Send me the next one 📸

Accepted Answer · 2026-05-11T14:19:42.647Z

Welcome back aborkugyaanku4@gmail.com — been a while! Let's pick up where you left off. To solve this problem, we first need to determine the components of the gravitational force acting on the particle parallel and perpendicular to the inclined plane, and then calculate the maximum static friction force. Given: Mass m = 10 kg Angle of inclination = 30^ Coefficient of static friction = 0.4 Acceleration due to gravity g = 10 m/s^2 Step 1: Calculate the components of the gravitational force and the normal force. The weight of the particle is W = mg. W = 10 kg × 10 m/s^2 = 100 N The component of the weight parallel to the plane (acting down the plane) is W_. W_ = W = 100 N × 30^ = 100 N × 0.5 = 50 N The component of the weight perpendicular to the plane is W_. W_ = W = 100 N × 30^ = 100 N × sqrt(3)2 ≈ 100 N × 0.866 = 86.6 N The normal force N is equal to the perpendicular component of the weight. N = W_ = 86.6 N The maximum static friction force f_s is given by N. f_s = N = 0.4 × 86.6 N = 34.64 N i) To hold the particle at rest: The particle tends to slide down the plane due to W_ = 50 N. Since W_ (50 N) is greater than the maximum static friction force f_s (34.64 N), an external force F_1 must be applied up the plane to prevent it from sliding down. In this case, the friction force acts up the plane, opposing the downward motion. Step 2: Calculate the force F_1 required to hold the particle at rest. For equilibrium, the sum of forces parallel to the plane must be zero. F_1 + f_s = W_ F_1 = W_ - f_s F_1 = 50 N - 34.64 N F_1 = 15.36 N The force required to hold the particle at rest is 15.36 N. ii) To just make the particle move up the plane: An external force F_2 must be applied up the plane. In this case, the particle is about to move up, so the friction force f_s acts down the plane, opposing the upward motion. The component of gravity W_ also acts down the plane. Step 3: Calculate the force F_2 required to just make the particle move up the plane. For impending upward motion, the sum of forces parallel to the plane must be zero. F_2 = W_ + f_s F_2 = 50 N + 34.64 N F_2 = 84.64 N The force required to just make the particle move up the plane is 84.64 N. Send me the next one 📸

A particle of mass 10 kg is placed on an inclined plane at an angle of 30 degrees. The coefficient of static friction is 0.4. Determine the forces involved and whether the particle remains at rest.

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A particle of mass 10 kg is placed on an inclined plane at an angle of 30 degrees. The coefficient of static friction is 0.4. Determine the forces involved and whether the particle remains at rest.

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