Solution Found!
Answer: A garage door is mounted on an overhead rail (Fig.
Chapter 11, Problem 83P(choose chapter or problem)
A garage door is mounted on an overhead rail (Fig. P11.83). The wheels at \(A\) and\(B\) have rusted so that they do not roll, but rather slide along the track. The coefficient of kinetic friction is 0.52. The distance between the wheels is \(2.00 \mathrm{~m}\), and each is \(0.50 \mathrm{~m}\) from the vertical sides of the door. The door is uniform and weighs 950 N. It is pushed to the left at constant speed by a horizontal force \(\vec{F}\)
(a) If the distance \(h\) is \(1.60 \mathrm{~m}\), what is the vertical component of the force exerted on each wheel by the track?
(b) Find the maximum value h can have without causing one wheel to leave the track.
Equation Transcription:
Text Transcription:
A
B
2.00 m
0.50 m
Vec F
h
1.60 m
2.00 m
3.00 m
Questions & Answers
QUESTION:
A garage door is mounted on an overhead rail (Fig. P11.83). The wheels at \(A\) and\(B\) have rusted so that they do not roll, but rather slide along the track. The coefficient of kinetic friction is 0.52. The distance between the wheels is \(2.00 \mathrm{~m}\), and each is \(0.50 \mathrm{~m}\) from the vertical sides of the door. The door is uniform and weighs 950 N. It is pushed to the left at constant speed by a horizontal force \(\vec{F}\)
(a) If the distance \(h\) is \(1.60 \mathrm{~m}\), what is the vertical component of the force exerted on each wheel by the track?
(b) Find the maximum value h can have without causing one wheel to leave the track.
Equation Transcription:
Text Transcription:
A
B
2.00 m
0.50 m
Vec F
h
1.60 m
2.00 m
3.00 m
ANSWER:
Solution 83P
Step 1:
a) We have to calculate the force exerted by each wheel on the door in this question. Since the door is moving horizontally in a constant speed, we can tell that, there is no horizontal force is acting on the door.
That is, the frictional force acting opposite to the applied force = Applied force from left direction
Given that, the applied force is F
Then, the frictional force, f = μN
Where, μ - coefficient of kinetic friction
N - normal force due to the door.
Provided, the normal force, N = 950 N and μ = 0.52
Therefore, f = 0.52 × 950 N = 494 N
Therefore, the force applied, F = 494 N