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 use the force components involved.
For the horizontal direction:
The force pushing the box along the wall is the horizontal component of the girl's force, which can be calculated using trigonometry:
Horizontal force = force * cos(angle) = 50 N * cos(45°) = 50 N * (√2/2) = 50√2/2 N = 25√2 N
For the vertical direction:
The weight of the box is the only force acting vertically. Therefore, the vertical force is the weight of the box, which is 40 N.
Since the box is lowered down at constant speed, the force of kinetic friction must be equal to the horizontal force component. So we have:
Force of kinetic friction = 25√2 N
Now, we can calculate the coefficient of kinetic friction using the equation:
Coefficient of kinetic friction = Force of kinetic friction / Normal force
The normal force is equal to the weight of the box, which is 40 N. Therefore:
Coefficient of kinetic friction = (25√2 N) / (40 N) = 0.625
So, the coefficient of kinetic friction between the box and the wall is 0.625.
b) Now, let's determine the range of force (Fmin, Fmax) the girl can push the stationary box without causing it to move. In this case, we are dealing with the coefficient of static friction.
The maximum force the girl can apply without causing the box to move is given by:
Fmax = μs * Normal force
Here, μs is the coefficient of static friction, which is 0.60, and the Normal force is again equal to the weight of the box, which is 40 N.
Therefore:
Fmax = (0.60) * (40 N) = 24 N
To determine the minimum force required to start the box moving, we consider that the minimum force required is equal to the force of static friction:
Fmin = Force of static friction = μs * Normal force = (0.60) * (40 N) = 24 N
So, the range of force (Fmin, Fmax) the girl can push the box without causing it to move is (24 N, 24 N).