The coefficient of static friction between the m = 3.75-kg crate and the 35.0° incline of the figure below is 0.320. What minimum force vector F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
gravity force down the plane=mg*cosTheta
friction force=mu(mg*sinTheta+F)
set them equal, solve for F
To determine the minimum force required to prevent the crate from sliding down the incline, we need to balance the forces along the incline. There are two main forces acting on the crate:
1. The gravitational force, which is the weight of the crate acting vertically downward. Its magnitude can be calculated as:
F_gravity = m * g
where m is the mass of the crate (3.75 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F_gravity = 3.75 kg * 9.8 m/s^2 = 36.75 N
2. The force of static friction, which acts parallel to the incline and prevents the crate from sliding. The maximum value of static friction is given by:
F_friction_max = coefficient of static friction * normal force
In this case, the normal force is the component of the weight perpendicular to the incline. It can be calculated as:
F_normal = F_gravity * cos(theta),
where theta is the angle of the incline (35 degrees in this case).
F_normal = 36.75 N * cos(35°) ≈ 30.06 N
Therefore, the maximum static friction force is:
F_friction_max = 0.320 * 30.06 N ≈ 9.62 N
To prevent the crate from sliding down the incline, the applied force perpendicular to the incline (F) must be equal to the force of static friction. So:
F = F_friction_max = 9.62 N
Therefore, a minimum force of 9.62 N must be applied perpendicular to the incline to prevent the crate from sliding down.