A box with a mass of 5.0 kg is placed on a flat sloping surface elevated at 40o to the horizontal. The coefficient of static friction is 0.76. It is prevented from slipping down the slope by a force applied parallel to the slope. What is the magnitude of this force?
To find the magnitude of the force applied parallel to the slope that prevents the box from slipping down, we need to consider the forces acting on the box.
First, let's calculate the gravitational force acting on the box. The gravitational force can be found using the formula: F_gravity = m * g, where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given that the mass of the box is 5.0 kg, the gravitational force will be:
F_gravity = 5.0 kg * 9.8 m/s^2 = 49 N
Next, let's calculate the maximum frictional force that can act on the box. The maximum static frictional force can be determined using the formula: F_friction_max = coefficient_of_static_friction * F_normal, where the coefficient_of_static_friction is given as 0.76.
The normal force acting on the box is the component of the gravitational force perpendicular to the slope. It can be calculated as: F_normal = m * g * cos(theta), where theta is the angle of the slope (40 degrees).
F_normal = 5.0 kg * 9.8 m/s^2 * cos(40 degrees) = 37.4963 N
Now we can find the maximum static frictional force:
F_friction_max = 0.76 * 37.4963 N = 28.4951 N
The force applied parallel to the slope must be equal to or greater than the maximum static frictional force to prevent the box from slipping down. Therefore, the magnitude of the force applied parallel to the slope is 28.4951 N.