Our 10kg mystery box rests on a horizontal floor. The
coefficient of static friction is 0.4 and the coefficient
of kinetic friction is 0.3. Determine the force of
friction acting on the box if a horizontal external
applied force is exerted on it of magnitude:
• (a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N.
7 8 9 10
Student
To determine the force of friction acting on the box for different magnitudes of the applied force, we need to compare the respective forces to the maximum static friction force and the kinetic friction force.
Given data:
Mass of the box (m) = 10 kg
Coefficient of static friction (μs) = 0.4
Coefficient of kinetic friction (μk) = 0.3
(a) When the applied force is 0 (a = 0):
Since there is no applied force, the box remains at rest, and the force of friction is equal to the maximum static friction force:
Maximum static friction force (Fsf) = μs * Normal force
Normal force (N) = m * g
where g is gravitational acceleration (approximately 9.8 m/s^2).
Substituting the given values:
N = 10 kg * 9.8 m/s^2 = 98 N
Fsf = 0.4 * 98 N = 39.2 N
Therefore, when the applied force is 0 N, the force of friction acting on the box is 39.2 N.
(b) When the applied force is 10 N (a = 10 N):
Since the applied force is less than the maximum static friction force, the box remains at rest, and the force of friction is equal to the applied force:
Force of friction (Ff) = 10 N
Therefore, when the applied force is 10 N, the force of friction acting on the box is 10 N.
(c) When the applied force is 20 N (a = 20 N):
Since the applied force is greater than the maximum static friction force, the box starts moving. The force of friction will now be the kinetic friction force:
Kinetic friction force (Ffk) = μk * Normal force
Substituting the values:
Ffk = 0.3 * 98 N = 29.4 N
Therefore, when the applied force is 20 N, the force of friction acting on the box is 29.4 N.
(d) When the applied force is 38 N (a = 38 N):
Since the applied force is still greater than the maximum static friction force, the box continues to move, and the force of friction remains the same as the kinetic friction force:
Force of friction (Ff) = 29.4 N
Therefore, when the applied force is 38 N, the force of friction acting on the box is 29.4 N.
(e) When the applied force is 40 N (a = 40 N):
Since the applied force is now equal to the maximum static friction force, the box is on the verge of transitioning from static to kinetic friction. The maximum static friction force is overcome, and the force of friction becomes equal to the kinetic friction force:
Force of friction (Ff) = 29.4 N
Therefore, when the applied force is 40 N, the force of friction acting on the box is 29.4 N.
To determine the force of friction acting on the box for different magnitudes of external applied force, we need to consider the cases where the box is at rest (static friction) and when it is in motion (kinetic friction).
Let's go through each case step-by-step:
(a) When the external applied force is 0 N:
Since there is no external force, the box remains at rest. In this case, the force of friction is equal to the maximum static friction, which can be calculated using the formula:
Static Friction = coefficient of static friction x Normal force
The normal force on the box is equal to the weight of the box, which can be calculated using the formula:
Weight = mass x gravitational acceleration (g)
So, for the given situation:
Static Friction = 0.4 x (mass x g)
Substituting the values, we get:
Static Friction = 0.4 x (10 kg x 9.8 m/s^2)
Static Friction = 39.2 N
Therefore, the force of friction when the external applied force is 0 N is 39.2 N.
(b) When the external applied force is 10 N:
In this case, the applied force is less than the maximum static friction, so the box remains at rest. Hence, the force of friction will be equal to the applied force. Therefore, the force of friction is 10 N.
(c) When the external applied force is 20 N:
When the applied force is greater than the maximum static friction, the box will start to move. At this point, the force of friction switches from static friction to kinetic friction.
The force of kinetic friction is given by:
Kinetic Friction = coefficient of kinetic friction x Normal force
So, for the given situation:
Kinetic Friction = 0.3 x (mass x g)
Substituting the values, we get:
Kinetic Friction = 0.3 x (10 kg x 9.8 m/s^2)
Kinetic Friction = 29.4 N
Therefore, the force of friction when the external applied force is 20 N is 29.4 N.
(d) When the external applied force is 38 N:
In this case, the applied force is still greater than the maximum static friction, so the box will already be in motion. Hence, the force of friction will be the same as the kinetic friction, which is 29.4 N.
(e) When the external applied force is 40 N:
The applied force is greater than the maximum static friction, so the box continues to move with the same kinetic friction as in the previous case. Therefore, the force of friction remains 29.4 N.
To summarize:
(a) Force of Friction = 39.2 N
(b) Force of Friction = 10 N
(c) Force of Friction = 29.4 N
(d) Force of Friction = 29.4 N
(e) Force of Friction = 29.4 N