A loaded penguin sled weighing 60 N rests on a plane inclined at 20° to the horizontal. Between the sled and the plane the coefficient of static friction is 0.27, and the coefficient of kinetic friction is 0.13.
(a) What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane?
N
(b) What is the minimum magnitude F that will start the sled moving up the plane?
N
(c) What value of F is required to move the block up the plane at constant velocity?
N
(a) Well, I don't think penguins are big fans of slipping down planes, so we need to find the minimum force to prevent that from happening. We can start by calculating the force of gravity acting down the incline. The force of gravity is given by:
Fg = m * g
Where m is the mass of the sled and g is the acceleration due to gravity. Since weight = mass * gravity, we can rewrite this as:
Fg = weight = 60 N
Now, we need to find the component of this force parallel to the plane. This is given by:
Fpar = Fg * sin(θ)
= 60 N * sin(20°)
Next, we need to find the maximum force of static friction that can act to prevent the sled from slipping. This is given by:
Fsf_max = μs * N
Where μs is the coefficient of static friction and N is the normal force. The normal force can be found from:
N = Fg * cos(θ)
= 60 N * cos(20°)
Now we can calculate the minimum force required to prevent slipping:
F_min = Fpar + Fsf_max
= 60 N * sin(20°) + 0.27 * 60 N * cos(20°)
(b) Now, let's find the minimum force required to start the sled moving up the plane. In this case, we are dealing with kinetic friction. The maximum force of kinetic friction is given by:
Fkf_max = μk * N
Where μk is the coefficient of kinetic friction. So, the minimum force required to start the sled moving is:
F_start = Fpar + Fkf_max
= 60 N * sin(20°) + 0.13 * 60 N * cos(20°)
(c) Finally, to keep the sled moving at a constant velocity, the force of kinetic friction must be equal to the applied force F. So, we have:
F = Fpar + Fkf_max
= 60 N * sin(20°) + 0.13 * 60 N * cos(20°)
I hope that helps!
To solve these problems, we can use the concept of forces and the laws of motion. Let's break down each part step-by-step:
(a) What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane?
To prevent the sled from slipping down the plane, the force of static friction, Fs, must oppose the component of force parallel to the plane, Fpar.
The force of static friction (Fs) can be calculated using the formula:
Fs ≤ μs * N
Where Fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.
The normal force (N) can be calculated by finding the component of weight perpendicular to the plane:
N = mg * cos(θ)
Where m is the mass of the sled, g is the acceleration due to gravity, and θ is the angle of inclination.
Given:
Weight of sled, W = 60 N
θ = 20°
μs = 0.27
First, calculate the normal force (N):
N = 60 N * cos(20°)
Next, calculate the force of static friction (Fs):
Fs = 0.27 * N
Finally, find the minimum magnitude of the force F (Fpar) needed to prevent the sled from slipping down the plane:
F ≥ Fs
This gives the answer to part (a).
(b) What is the minimum magnitude F that will start the sled moving up the plane?
To start the sled moving up the plane, the force of static friction must be equal to the force parallel to the plane, Fpar.
Using the same calculations as in part (a), find the force of static friction (Fs). This Fs will be the minimum magnitude of force needed to start the sled moving up the plane.
This gives the answer to part (b).
(c) What value of F is required to move the sled up the plane at constant velocity?
To move the sled up the plane at constant velocity, the force of dynamic or kinetic friction (Fk) must be equal to the force parallel to the plane, Fpar.
The force of dynamic or kinetic friction can be calculated using the formula:
Fk = μk * N
Where Fk is the force of dynamic or kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force.
Using the same calculations as in part (a), find the normal force (N) and calculate the force of dynamic or kinetic friction (Fk).
Finally, find the value of force F (Fpar) required to move the sled up the plane at constant velocity:
F = Fk
This gives the answer to part (c).
To solve this problem, we need to consider the forces acting on the sled and apply Newton's laws of motion. Let's break it down step by step.
First, let's draw a free body diagram to visualize the forces acting on the sled. We have the weight of the sled (60 N) acting vertically downwards and the normal force (N) acting perpendicular to the plane. There are two frictional forces to consider: the static frictional force (fs) and the kinetic frictional force (fk). We need to determine how these forces are related to the applied force (F).
Now, let's answer each part of the question:
(a) To prevent the sled from slipping down the plane, we need to find the minimum force required to counteract the force of friction.
The force of friction can be calculated as fs = coefficient of static friction x normal force.
fs = 0.27 x N
Since the sled is at the verge of slipping, the force of static friction will be at its maximum value, which is equal to the coefficient of static friction times the normal force.
At the verge of slipping, the force parallel to the plane (F) must balance the force of static friction to prevent any motion. Therefore, F = fs.
Substituting the value of the force of static friction, we have:
F = 0.27 x N
(b) To start the sled moving up the plane, we need to overcome the maximum force of static friction.
The force required to start the sled moving up the plane is equal to the force of static friction at its maximum value. So, F = fs.
Substituting the value of the force of static friction, we have:
F = 0.27 x N
(c) To move the block up the plane at a constant velocity, we need to overcome the force of kinetic friction.
The force of kinetic friction can be calculated as fk = coefficient of kinetic friction x normal force.
fk = 0.13 x N
To move the block at a constant velocity, the force applied (F) must balance the force of kinetic friction. Therefore, F = fk.
Substituting the value of the force of kinetic friction, we have:
F = 0.13 x N
Please note that the values of normal force (N) and coefficients of friction used in these equations are not provided in the question. You'll need to determine them or find additional information to solve the problem completely.
F = 60N @ 20o. = Force of the sled.
Fp = 60*sin20 = 20.52 N. = Force parallel to the plane.
Fv = 60*cos20 = 56.38 N. = Force perpendicular to the plane.
Fs = u*Fv = 0.27*56.38 = 15.22 N. = Force of static friction.
Fk = u*Fv = 0.13*56.38 = 7.33 N. = Force of kinetic friction.
a. Fap-Fp-Fs = m*a = m*0
Fap = Fp + Fs = 20.52 + 15.22 = 35.74 N.
= Force applied.
b. Fap-Fp-Fs = m*a = m*0 = 0
Fap = Fp+Fs = 20.52+15.22 = 35.74 N.
c. Fap-Fp-Fk = m*a = m*0 = 0
Fap = Fp+Fk = 20.52+7.33 = 27.85 N.