A 1000N crate is being pushed across a level floor at a constant speed by a force F of 300N at an angle of 20.0 degrees. (a) What is the coefficient of kinetic friction between the crate and the floor? (b) If the 300-N force is instead pulling the block at an angle of 20.0 degrees above the horizontal, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in (a).
how is the angle measured?
N=mg±Fsinα
+ if α is measured below the horizontal
- if α is measured above the horizontal
(a) mgcosα =μN
μ = mgcosα/N
(b) ma=Fcosα –F(fr) =
= Fcosα – μN
Solve for ‘a’
To find the coefficient of kinetic friction between the crate and the floor, we need to analyze the forces acting on the crate.
(a) First, let's consider the forces acting horizontally. The force pushing the crate, denoted as F_push, has a magnitude of 300N and is at an angle of 20.0 degrees. The force of kinetic friction, denoted as F_friction, is acting in the opposite direction. Since the crate is moving at a constant speed, the net force horizontally is zero.
The formula to calculate the force of kinetic friction is given by:
F_friction = coefficient_of_friction * normal_force
where the normal force is equal to the weight of the crate, which is the force acting vertically downwards. In this case, the normal force is equal to the weight of the crate:
Normal force = Weight = m * g
where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s²).
Since the net force horizontally is zero, we have:
F_push - F_friction = 0
Substituting the values we know:
300N - F_friction = 0
F_friction = 300N
Therefore, the coefficient of kinetic friction can be calculated as:
coefficient_of_friction = F_friction / normal_force
Now we need to calculate the normal force:
Normal force = Weight = m * g
To calculate the mass of the crate, we can use Newton's second law:
Sum of forces vertically = 0
In this case, there are two forces acting vertically: the weight downwards and the vertical component of the pushing force upwards. Since the crate is not accelerating vertically, the sum of the forces vertically is zero:
F_push_vertical - Weight = 0
F_push_vertical = F_push * sin(angle)
Setting up the equation:
F_push_vertical = F_push * sin(angle) = 300N * sin(20.0°)
Now we have:
F_push_vertical - Weight = 0
F_push * sin(angle) - m * g = 0
Solving for mass:
m = (F_push * sin(angle)) / g
Substituting the given values:
m = (300N * sin(20.0°)) / 9.8 m/s²
Once you have the mass, you can calculate the normal force:
Normal force = Weight = m * g
After finding the normal force, you can calculate the coefficient of kinetic friction by dividing the force of kinetic friction by the normal force.
(b) To find the acceleration when the force is pulling the crate instead of pushing it, we can use Newton's second law:
Sum of forces horizontally = mass * acceleration
In this case, there are two forces acting horizontally: the force pulling the crate and the force of kinetic friction. However, the direction of the force of kinetic friction will be opposite to the force pulling the crate since it is trying to resist the motion.
Therefore, the equation becomes:
F_pull - F_friction = mass * acceleration
Substituting the values we know:
F_pull - F_friction = m * a
Since the force of kinetic friction is the same as in part (a), we can use the same coefficient of kinetic friction:
F_pull - coefficient_of_friction * Normal force = m * a
Now we can substitute the values we've already calculated:
F_pull - coefficient_of_friction * normal_force = m * a
Substituting the values we've already calculated for the coefficient of friction and normal force, we can find the acceleration a.