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Answer :
Final answer:
To find the magnitude of the minimum force the person needs to apply to keep the box stationary on the inclined ramp, we can use the equation F_friction = μ_s N, where μ_s is the coefficient of static friction and N is the normal force. N can be calculated using the equation N = mg cos(theta), where m is the mass of the box, g is the acceleration due to gravity, and theta is the angle of the ramp.
Explanation:
To find the magnitude of the minimum force the person needs to apply to keep the box stationary, we need to consider the forces acting on the box. The force of gravity acting vertically downward can be resolved into two components: one perpendicular to the ramp and one parallel to the ramp.
The component of the weight parallel to the ramp is responsible for the force of friction. Since the box is at rest, the force of friction must be equal and opposite to the applied force. We can use the equation F_friction = μ_s N, where μ_s is the coefficient of static friction and N is the normal force, to find the minimum force needed to keep the box stationary. The normal force can be calculated using the equation N = mg cos(theta), where m is the mass of the box, g is the acceleration due to gravity, and theta is the angle of the ramp.
In this case, the mass of the box is given as M = 5.0 kg and the coefficient of static friction is given as 0.30. The angle of the ramp is given as 30°. The acceleration due to gravity is approximately 9.8 m/s^2. Plugging these values into the equations, we get: N = (5.0 kg)(9.8 m/s^2) cos(30°) = 42.5 N. And F_friction = (0.30)(42.5 N) = 12.8 N. Therefore, the magnitude of the minimum force the person needs to apply to keep the box stationary is approximately 12.8 N.
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Final answer:
The minimum horizontal force that needs to be applied to keep the box stationary on the inclined plane is approximately 37.23 N. However, amongst the options provided, the closest correct answer would be 36.6 N (choice E).
Explanation:
The question requires finding the minimum horizontal force that needs to be applied to prevent a box of mass M = 5.0 kg, lying on an inclined plane at an angle of 30° to the horizontal, from sliding down. The coefficient of static friction between the box and the incline is given as 0.30.
The weight of the box in the direction perpendicular to the inclined plane is W' = M*g*cos(θ), and along the inclined plane it's W'' = M*g*sin(θ). Here, M is the mass of the box, g is the acceleration due to gravity (9.8 m/s²) and θ is the inclined angle. Thus, W' = 5*9.8*cos(30°) = 42.44 N and W'' = 5*9.8*sin(30°) = 24.5 N.
The friction force, Ff acting along the incline is Ff=µ*N, where µ is the coefficient of static friction and N is the normal force. Therefore, Ff =0.30*42.44 N = 12.73 N.
To keep the box stationary, the minimum horizontal force someone needs to exert should counterbalance both the component of the box's weight acting down the incline and the friction force.
Hence, F = W'' + Ff = 24.5 N + 12.73 N = 37.23 N. The given options do not include this value, however, the closest correct answer provided would be (E) F = 36.6N.
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