A boy exerts a horizontal force of 111 N on a box with a mass of 32.7 kg.
HINT - (a) Make a free-body diagram and apply the x-component of Newton's second law.
(b) Apply the relationship between static friction and the normal force.
(a)
If the box doesn't move, what's the magnitude of the static friction force (in N)?
(b)
What is the minimum possible value of the coefficient of static friction between the box and the floor? (Assume the box remains stationary.)
(a) > 111 N
(b) m g μ = 111 N ... μ = 111 / (32.7 * 9.81)
(a) Well, if the box doesn't move, that means the static friction force must be equal in magnitude and opposite in direction to the horizontal force being applied. So the magnitude of the static friction force would also be 111 N!
(b) Ah, the coefficient of static friction. Well, if the box remains stationary, that means the static friction force is at its maximum and just balancing out the applied force. The relationship between static friction and the normal force is given by the equation Fs ≤ μs * Fn, where Fs is the static friction force, μs is the coefficient of static friction, and Fn is the normal force.
In this case, since the box is on a flat surface, the normal force would be equal in magnitude to the weight of the box, which is m * g. Therefore, the minimum possible value of the coefficient of static friction would be Fs max / Fn = 111 N / (32.7 kg * g), where g is the acceleration due to gravity.
(a)
To find the magnitude of the static friction force, let's apply the x-component of Newton's second law. The equation is:
ΣFx = max
Since the box is not moving, the acceleration (a) is zero. Therefore, the equation becomes:
ΣFx = 0
The only horizontal force acting on the box is the boy's exerted force of 111 N, in the positive x-direction. Therefore:
ΣFx = 111 N - Static friction force = 0
To find the static friction force, we rearrange the equation:
Static friction force = 111 N
So, the magnitude of the static friction force is 111 N.
(b)
To find the minimum possible value of the coefficient of static friction, we need to consider the relationship between static friction and the normal force. The formula is:
Static friction force = coefficient of static friction × normal force
From part (a), we found that the magnitude of the static friction force is 111 N. Since the box is on a horizontal surface, the normal force is equal to the weight of the box, which can be calculated using the formula:
Normal force = mass × gravitational acceleration
Given that the mass of the box is 32.7 kg and the gravitational acceleration is approximately 9.8 m/s^2, we can calculate the normal force as follows:
Normal force = 32.7 kg × 9.8 m/s^2 = 320.46 N (approximately)
Now, we can rearrange the formula for static friction force:
Coefficient of static friction = Static friction force / normal force
Substituting the values, we get:
Coefficient of static friction = 111 N / 320.46 N ≈ 0.346
Therefore, the minimum possible value of the coefficient of static friction between the box and the floor is approximately 0.346.
(a) To find the magnitude of the static friction force, we can use Newton's second law. The equation states that the net force in the x-direction is equal to the mass of an object multiplied by its acceleration. In this case, the box is not moving, so its acceleration is zero. Therefore, the net force in the x-direction is also zero.
The net force in the x-direction is the vector sum of the applied force and the static friction force, which acts in the opposite direction. Since the box is not moving, the magnitude of the static friction force is equal to the magnitude of the applied force, but in the opposite direction.
So, the magnitude of the static friction force is 111 N.
(b) The minimum possible value of the coefficient of static friction between the box and the floor can be found using the relationship between static friction and the normal force. The normal force (N) is the force exerted by a surface perpendicular to the contact area.
When the box is at rest, the normal force is equal to the weight of the box (mg), where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s²).
So, the normal force (N) = 32.7 kg * 9.8 m/s² = 320.46 N.
The static friction force (Fs) can be expressed as the product of the coefficient of static friction (μs) and the normal force:
Fs = μs * N.
Since the box is not moving, the static friction force (Fs) is equal to the magnitude of the applied force (111 N).
111 N = μs * 320.46 N.
Simplifying the equation, we find the coefficient of static friction (μs) = 111 N / 320.46 N.
So, the minimum possible value of the coefficient of static friction between the box and the floor is approximately 0.346.