A loaded penguin sled weighing 75 N rests on a plane inclined at angle θ = 24° to the horizontal (Fig. 6-28). Between the sled and the plane, the coefficient of static friction is 0.25, and the coefficient of kinetic friction is 0.13. (a) What is the minimum magnitude of the force , parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude F that will start the sled moving up the plane? (c) What value of F is required to move the sled up the plane at constant velocity?
Fs = (75N,24deg).
Fp=75sin24 = 30.51N. = Force parallel to plane acting down the plane.
Fv = 75cos24 = 68.52N. = Force perpen-
dicular to plane.
Ff = u*Fv = 0.25 * 68.52 = 17.13N. =
Force of static friction.
Fk = 0.13 * 68.52 = 8.91 = Force of
kinetic friction.
a. Fap - Fp - Fk = 0,
Fap=Fp + Ff = 30.51 + 8.91 = 39.42N =
Min Force applied.
b. Fap - Fp - Ff = 0,
Fap = Fp + Ff = 30.51 + 17.13 = 47.64N.
= Min force applied.
c. Fap = Fp + Fk = 30.51 + 8.91 = 39.42N.
part a is wrong
pt A: Fp - Ff. in this case, you would do 30.51 - 17.13 = 13.38
To solve this problem, we need to use the concepts of static and kinetic friction and the principles of Newton's laws of motion. Here's how we can approach each part of the problem:
(a) To find the minimum magnitude of the force parallel to the plane that will prevent the sled from slipping down, we need to calculate the maximum force of static friction. Static friction acts in the opposite direction of the impending motion.
The force of static friction can be calculated using the equation: fs ≤ μs * N
where fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force acting on the sled.
Let's break down the force components acting on the sled:
- The weight of the sled (mg) can be calculated as (75 N * 9.8 m/s^2) = 735 N.
- The normal force (N) acting perpendicular to the inclined plane can be calculated as N = mg * cos(θ), where θ is the angle of inclination. Substituting the values, we have N = (735 N) * cos(24°) = 670.17 N.
Now, substitute the values into the equation for static friction: fs ≤ (0.25) * (670.17 N)
Therefore, the maximum force of static friction is fs ≤ 167.54 N.
Hence, the minimum magnitude of the force parallel to the plane required to prevent the sled from slipping down is 167.54 N.
(b) To find the minimum magnitude of force, F, required to start the sled moving up the plane, we need to consider the force of static friction again. In this case, we need to overcome the static frictional force to initiate motion.
Since the sled is about to start moving, the force of static friction reaches its maximum value, which we calculated in part (a) as 167.54 N.
Therefore, the minimum magnitude of the force, F, required to start the sled moving up the plane is also 167.54 N.
(c) To find the minimum value of force, F, required to move the sled up the plane at constant velocity, we need to consider the force of kinetic friction. Once the sled is in motion, the force of kinetic friction will come into play.
The force of kinetic friction can be calculated using the equation: fk = μk * N
where fk is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force acting on the sled.
Using the same weight and normal force values as before, we can calculate the force of kinetic friction: fk = (0.13) * (670.17 N) = 87.13 N.
To move the sled up the plane at constant velocity, the applied force, F, must overcome the force of kinetic friction. So, the minimum value of force, F, required in this case is 87.13 N.
To summarize:
(a) The minimum magnitude of the force parallel to the plane that will prevent the sled from slipping down is 167.54 N.
(b) The minimum magnitude of force, F, required to start the sled moving up the plane is also 167.54 N.
(c) The minimum value of force, F, required to move the sled up the plane at constant velocity is 87.13 N.