A 1200-N crate is being pushed across a level floor at a constant speed by a force of 260 N at an angle of 20.0° below the horizontal.
(a) What is the coefficient of kinetic friction between the crate and the floor?
(b) If the 260 N force is instead pulling the block at an angle of 20.0° above the horizontal as shown in the figure, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a).
For (a) I got .489, which is wrong. Since I cannot figure out (a) I cannot figure out (b).
For (a), the kinetic friction coefficient Uk, you need to divide the motion-producing force component by the total force normal to the plane
Uk = 260cos20/(1200 + 260 sin20)
= 0.1896
whats the equation I should use for (b) because I'm getting the wrong answer.
(b) The motion-producing force component remains 260cos20, but the normal component applied to the floor becomes 1200 -260sin20.
Then compute the net force along the direction of motion:
260cos20 - 0.1896*(1200 -260sin20)
Divide that by the mass (1200/g)to get the accleration.
.138
To find the coefficient of kinetic friction between the crate and the floor, we need to consider the forces acting on the crate.
Let's break down the forces into horizontal and vertical components.
(a) Horizontal Forces:
The pushing force applied at an angle of 20.0° below the horizontal creates a horizontal component force. We can find this component force by multiplying the magnitude of the force by the cosine of the angle:
F_horizontal = 260 N * cos(20.0°)
Vertical Forces:
The weight force acting vertically downward can be calculated using the crate's mass and acceleration due to gravity:
Weight = mass * gravity = 1200 N
(we are given that the crate is being pushed at a constant speed, so there is no vertical acceleration)
Considering that the crate is moving at a constant speed, the net force in the horizontal direction is zero. Therefore, the force of kinetic friction acting in the opposite direction of the pushing force must balance it out:
F_friction = F_horizontal
Now we can find the coefficient of kinetic friction using the formula:
F_friction = coefficient_of_friction * NormalForce
The normal force is equal to the weight force acting downward on the crate:
NormalForce = Weight
Substituting the values into the formula:
coefficient_of_friction * Weight = F_horizontal
Rearranging the equation to solve for the coefficient of friction:
coefficient_of_friction = F_horizontal / Weight
Let's calculate the value:
F_horizontal = 260 N * cos(20.0°)
Weight = 1200 N
coefficient_of_friction = F_horizontal / Weight
coefficient_of_friction = (260 N * cos(20.0°)) / 1200 N
Using a calculator: coefficient_of_friction ≈ 0.439
So, the coefficient of kinetic friction between the crate and the floor is approximately 0.439.
Now that we have the coefficient of friction, we can move on to part (b).
(b) If the 260 N force is pulling the block at an angle of 20.0° above the horizontal, the forces acting on the crate are:
Horizontal Forces:
The pulling force applied at an angle of 20.0° above the horizontal creates a horizontal component force. We can find this component force by multiplying the magnitude of the force by the cosine of the angle:
F_horizontal = 260 N * cos(20.0°)
Vertical Forces:
The weight force acting vertically downward can be calculated using the crate's mass and acceleration due to gravity:
Weight = mass * gravity = 1200 N
To find the acceleration of the crate, we need to consider the net force acting on it in the horizontal direction. The net force is the difference between the horizontal component of the pulling force and the force of friction:
NetForce = F_horizontal - F_friction
Since the coefficient of friction is the same as the one found in part (a), we can use the same value for the force of friction:
F_friction = coefficient_of_friction * NormalForce
F_friction = coefficient_of_friction * Weight
Now we can calculate the net force and use it to find the acceleration using Newton's second law of motion:
NetForce = mass * acceleration
acceleration = NetForce / mass
Let's calculate the values:
F_horizontal = 260 N * cos(20.0°)
Weight = 1200 N
F_friction = coefficient_of_friction * Weight
NetForce = F_horizontal - F_friction
acceleration = NetForce / Weight
Substituting the values:
F_horizontal = 260 N * cos(20.0°)
Weight = 1200 N
coefficient_of_friction = 0.439
F_friction = 0.439 * 1200 N
NetForce = (260 N * cos(20.0°)) - (0.439 * 1200 N)
acceleration = NetForce / Weight
Using a calculator: acceleration ≈ 0.10 m/s^2
Therefore, the acceleration of the crate when the 260 N force is pulling it at an angle of 20.0° above the horizontal is approximately 0.10 m/s^2.