Finding Acceleration and Stopping Time in a Multi-Block Lifting System

In summary, this conversation discusses a lifting system modeled as two blocks connected by ropes to a third block lying on a horizontal surface. The acceleration of the blocks is found by considering the forces acting on each block. The difference in mass between m1 and m2 is limited to 100kg for the system to remain at rest, taking into account static friction with a coefficient of μs = 1.00. When considering kinetic friction with a coefficient of μk = 0.700, the system takes 0.292 seconds to stop if both m1 and m2 have a mass of 100kg.
  • #1
ChrisBrandsborg
120
1

Homework Statement


A lifting system can be modeled as two blocks of mass m1 and m2 hanging from two pulleys, connected by ropes to a third block of mass M = 100 kg, lying on top of a horizontal surface. There is gravity g = 9.80 m/s2.

a) Ignoring at first friction, find an expression for the acceleration of the blocks, as a function of the masses and gravity.

b) Now, we assume that the system starts at rest, and that there is static friction between the block M and the surface it is lying on. How much diffe- rence in mass can m1 and m2 have, for the system to not start moving, if the coefficient of static friction is μs = 1.00? Provide an expression as well as a numerical value.

c) Next, we assume that the system is moving with an initial speed vi = 2.00 m/s, when suddenly kinetic friction between M and the surface is triggered (the brakes are switched on). The coefficient of kinetic friction is μk = 0.700. How long does it take for the system to stop if m2 = m1 = 100 kg ? Provide an expression as well as a numerical value.

Homework Equations


ΣF = ma
F* = F

(Newton´s laws)

The Attempt at a Solution


I think I did a correctly, but I have some problems solving b.
I put positive direction downwards to the left.

a) Firstly I separated the system into 3 systems:

1) am1 = m1g - T1
2) am2 = T2 - m2g
3) This one is the third block, lying on the top of the surface.

am3 = T1 - T2

then add together:

so that: a (m1 + m2 + m3) = m1g - T1 + T2 -m2g + T1 - T2

a = (m1 + m2)*g / (m1 + m2 + m3)

b)
I know that μs = 1.00.
Fs = nμs = mgμs = 100kg * 9.80 * 1.00 = 980N

That means that m1g - m2g = 980N
Then divide by g:

m1 - m2 = 100kg

Is that the answer? That the difference in mass has to be less or equal to 100kg?
 
Last edited:
Physics news on Phys.org
  • #2
b)
I know that μs = 1.00.
Fs = nμs = mgμs = 100kg * 9.80 * 1.00 = 980N

That means that m1g - m2g = 980N
Then divide by g:

m1 - m2 = 100kg

Is that the answer? That the difference in mass has to be less or equal to 100kg?
 
  • #3
I tried c as well:

vi = 2.00 m/s
μk = 0.700
How long does it take for the system to stop?

Fk = ma
μkmg = ma

a = μkg

v = vi + μkgt

t = vikg

t = 2.00/0.700*9.80 = 0.292s

Correct? Or do I have to use the masses as well? m1 = m2 = 100kg?
 

1. How does the braking system of an elevator work?

The braking system of an elevator typically uses a combination of mechanical and electrical components. When the elevator car reaches its designated floor, the brakes are engaged by an electric motor, which presses a friction pad against the guide rail to slow and stop the car.

2. What are the potential risks of an elevator braking failure?

If the braking system of an elevator fails, the elevator car could continue to move uncontrollably, potentially resulting in a collision with the top or bottom of the elevator shaft. This could also cause injury to passengers inside the car.

3. How often should elevator brakes be inspected?

Elevator brakes should be inspected at least once a year by a certified elevator technician. However, if the elevator experiences heavy use, it may be necessary to inspect the brakes more frequently.

4. Can extreme weather conditions affect the braking system of an elevator?

Extreme weather conditions, such as extreme heat or cold, can affect the performance of elevator brakes. Extreme heat can cause the brake pads to wear down more quickly, while extreme cold can cause the brake fluid to thicken and decrease its effectiveness.

5. How can elevator braking systems be improved?

Many modern elevators use regenerative braking systems, which convert the kinetic energy of the descending elevator car into electricity to power the elevator's motor. This not only improves energy efficiency but also reduces wear and tear on the braking system.

Similar threads

  • Introductory Physics Homework Help
Replies
23
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
992
  • Introductory Physics Homework Help
Replies
19
Views
11K
  • Introductory Physics Homework Help
Replies
3
Views
963
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
2
Replies
64
Views
12K
Back
Top