A box of textbooks of mass 25.3 kg rests on a loading ramp that makes an angle (ALFA) with the horizontal. The coefficient of kinetic friction is 0.260 and the coefficient of static friction is 0.330 .
A) As the angle (ALFA) is increased, find the minimum angle at which the box starts to slip
B) At this angle, find the acceleration once the box has begun to move
C)At this angle, how fast will the box be moving after it has slid a distance 4.60m along the loading ramp?
A) Use the static friction coefficient for this one. Call it Us. When the maximum friction coefficient equals the downhill component of the weight, slipping begins. Call the angle A
M g sin A = M g cos A * Us
Us = tan A
A = arctan (0.33) = 18.3 degrees
B) Once slipping starts, the friction coefficient is lowered to Uk. The net force down the ramp is
F = M g sin A - Mg cos A * Uk
and the acceleration is F/M
Solve for the acceleration, a.
C) W = sqrt(2 a X), where X is the distance moved. Use the 'a' from part B.
Thank you very much !
Nangsung.
Why do you multiply it by the distance by 2?
You're welcome! Here are the step-by-step solutions to your questions:
A) To find the minimum angle at which the box starts to slip, we can use the static friction coefficient. When the maximum static friction coefficient equals the downhill component of the weight, slipping begins. The maximum static friction coefficient can be calculated using the equation:
Us = tan(A),
where Us is the static friction coefficient and A is the angle.
Given that Us = 0.33, we can solve for A:
A = arctan(0.33) = 18.3 degrees.
Therefore, the minimum angle at which the box starts to slip is 18.3 degrees.
B) Once slipping starts, the friction coefficient is lowered to the kinetic friction coefficient. The net force down the ramp can be calculated using the equation:
F = Mg * sin(A) - Mg * cos(A) * Uk,
where F is the net force, Mg is the weight of the box, A is the angle, and Uk is the kinetic friction coefficient.
To find the acceleration, we divide the net force by the mass of the box:
a = F / M.
C) To find the speed of the box after sliding a distance of 4.60m along the loading ramp, we can use the equation:
W = sqrt(2 * a * X),
where W is the final velocity, a is the acceleration from part B, and X is the distance moved.
I hope this helps! Let me know if you have any further questions.
A) To find the minimum angle (ALFA) at which the box starts to slip, we need to consider the maximum friction force that can be exerted on the box before it starts to move. This maximum force of friction is given by the equation:
Friction force (Ff) = coefficient of static friction (Us) * normal force
The normal force is the force exerted perpendicular to the ramp, which is equal to the weight of the box (mass * gravitational acceleration). So, the equation becomes:
Ff = Us * mass * gravitational acceleration
In this case, the coefficient of static friction (Us) is given as 0.330, and the mass is given as 25.3 kg. The gravitational acceleration is approximately 9.81 m/s^2.
Substituting the values into the equation:
Ff = 0.330 * 25.3 kg * 9.81 m/s^2
Simplifying, we find:
Ff = 81.3789 N
Now, we can consider the forces acting on the box. The force down the ramp is given by:
Force down the ramp = mass * gravitational acceleration * sin(ALFA)
The force perpendicular to the ramp is given by:
Force perpendicular to the ramp = mass * gravitational acceleration * cos(ALFA)
For slipping to start, the force down the ramp (mg * sin(ALFA)) should be equal to the maximum friction force (Ff). So, we can set up the equation:
mass * gravitational acceleration * sin(ALFA) = Ff
Substituting the values:
25.3 kg * 9.81 m/s^2 * sin(ALFA) = 81.3789 N
Rearranging the equation to solve for the angle (ALFA):
sin(ALFA) = 81.3789 N / (25.3 kg * 9.81 m/s^2)
ALFA = arcsin(81.3789 N / (25.3 kg * 9.81 m/s^2))
ALFA ≈ 18.3 degrees
Therefore, the minimum angle at which the box starts to slip is approximately 18.3 degrees.
B) Once the box starts to slip, the friction force changes to the kinetic friction force. The equation for the kinetic friction force (Fk) is given by:
Fk = coefficient of kinetic friction (Uk) * normal force
In this case, the coefficient of kinetic friction (Uk) is given as 0.260.
The net force acting on the box is given by the force down the ramp minus the force opposing the motion (Fk), which is:
Net force = mass * gravitational acceleration * sin(ALFA) - Uk * mass * gravitational acceleration * cos(ALFA)
The acceleration of the box is given by:
Acceleration (a) = Net force / mass
Substituting the values:
a = (mass * gravitational acceleration * sin(ALFA) - Uk * mass * gravitational acceleration * cos(ALFA)) / mass
Simplifying the equation:
a = gravitational acceleration * (sin(ALFA) - Uk * cos(ALFA))
Using ALFA ≈ 18.3 degrees and Uk = 0.260, we can calculate the acceleration (a) in meters per second squared.
C) To find the speed at which the box will be moving after it has slid a distance of 4.60 meters along the loading ramp, we can use the equation:
Final velocity (Vf) = sqrt(2 * acceleration * distance)
Substituting the values:
Vf = sqrt(2 * a * 4.60)
Using the value of the acceleration (a) calculated in part B, we can find the final velocity (Vf) in meters per second.
Please note that the above calculations assume ideal conditions and neglect any other factors such as air resistance.