A box is being lowered down a wall at constant speed by a girl, as shown. Her hand is pushing on the edge of the box with 50 N of force at a 45◦ angle with respect to the horizontal. The box weighs 40 N. Find the coefficient of kinetic friction between the box and the wall.
b) Now imagine the box is stationary. The coefficient of static friction between the box and wall is 0.60. The girl still pushes at a 45◦ angle. Over what range of force (Fmin,Fmax) can the girl push the box without causing it to move?
To find the coefficient of kinetic friction between the box and the wall, we need to first find the net force acting on the box in the horizontal direction.
1) Net Force in the Horizontal Direction:
The force applied by the girl can be separated into horizontal and vertical components. The horizontal component of the force can be calculated using the angle of 45 degrees.
Horizontal Force = Force * cos(angle)
Horizontal Force = 50 N * cos(45°)
Horizontal Force = 50 N * 0.707
Horizontal Force = 35.35 N (approx)
The weight of the box is acting in the opposite direction, which is downward.
Weight of the box = 40 N (downward)
Net Horizontal Force = Horizontal Force - Weight of the box
Net Horizontal Force = 35.35 N - 40 N
Net Horizontal Force = -4.65 N
Since the box is being lowered down the wall at a constant speed, the net force in the horizontal direction is zero.
Net Horizontal Force = 0 N
Now, we can calculate the kinetic friction force:
Kinetic Friction Force = Coefficient of Kinetic Friction * Normal Force
Since the box is being lowered, the normal force is equal to the weight of the box.
Normal Force = Weight of the box
Normal Force = 40 N
Since the net horizontal force is zero, the kinetic friction force must be equal in magnitude to the applied force.
Kinetic Friction Force = 4.65 N
Therefore, the coefficient of kinetic friction can be calculated as:
Coefficient of Kinetic Friction = Kinetic Friction Force / Normal Force
Coefficient of Kinetic Friction = 4.65 N / 40 N
Coefficient of Kinetic Friction = 0.11625 (approximately)
So, the coefficient of kinetic friction between the box and the wall is approximately 0.11625.
b) To find the range of force (Fmin, Fmax) the girl can push the box without causing it to move when the box is stationary, we need to consider the static friction.
The maximum static friction force can be calculated as:
Maximum Static Friction Force = Coefficient of Static Friction * Normal Force
Maximum Static Friction Force = 0.60 * Normal Force
Maximum Static Friction Force = 0.60 * 40 N
Maximum Static Friction Force = 24 N
Since the box is stationary, the maximum static friction force must be equal in magnitude to the applied force.
Maximum Static Friction Force = Fmax - Fmin
24 N = Fmax - Fmin
So, the range of force (Fmin, Fmax) can be found by solving the equation:
Fmax - Fmin = 24 N
The specific values of Fmin and Fmax will depend on the specific situation and any additional information provided.
To find the coefficient of kinetic friction between the box and the wall, we need to analyze the forces involved.
1) First, let's break down the force applied by the girl into its horizontal and vertical components:
- The horizontal component of the force is given by F_horizontal = F * cos(theta), where F is the magnitude of the force (50 N) and theta is the angle with respect to the horizontal (45 degrees). So, F_horizontal = 50 N * cos(45 degrees) = 35.355 N.
- The vertical component of the force is given by F_vertical = F * sin(theta), where F is the magnitude of the force (50 N) and theta is the angle with respect to the horizontal (45 degrees). So, F_vertical = 50 N * sin(45 degrees) = 35.355 N.
2) Now, let's analyze the forces acting on the box when it is being lowered down the wall at a constant speed:
- The weight of the box acts vertically downward with a magnitude of 40 N.
- The vertical component of the force applied by the girl cancels out the weight of the box, preventing it from falling or rising.
- The horizontal component of the force applied by the girl overcomes the friction force opposing the motion.
3) Since the box is moving at a constant speed, the applied force by the girl must be equal in magnitude and opposite in direction to the friction force. Therefore, we can equate the horizontal component of the applied force (35.355 N) to the friction force (F_friction) and find the coefficient of kinetic friction (μ_k):
F_friction = μ_k * N,
where N is the normal force exerted by the wall on the box, which is equal to the weight of the box (40 N).
So, 35.355 N = μ_k * 40 N.
Solving for μ_k, we find: μ_k = 35.355 N / 40 N = 0.8839.
Therefore, the coefficient of kinetic friction between the box and the wall is approximately 0.8839.
To calculate the range of force (F_min, F_max) that the girl can push the stationary box without causing it to move, we need to consider the maximum static friction force.
1) The maximum static friction force (F_friction_max) can be calculated using the equation:
F_friction_max = μ_s * N,
where μ_s is the coefficient of static friction (0.60) and N is the normal force exerted by the wall on the box, which is equal to the weight of the box (40 N).
So, F_friction_max = 0.60 * 40 N = 24 N.
2) The girl can push the box without causing it to move as long as her applied force (F) is less than or equal to the maximum static friction force (F_friction_max). Therefore, the range of force is given by (F_min, F_max) = (0 N, 24 N).
In summary, the girl can push the stationary box without causing it to move if her applied force is within the range of (0 N, 24 N), based on the coefficient of static friction of 0.60.