Blocks on pulley on incline plane

In summary, the mass M1 can be found by calculating the forces acting on M2, including the force of gravity, the acceleration, and the tension. Using the formula F=ma and setting the acceleration to be 0 in the y direction, the value of N can be found to be equal to mgcosθ. The final calculation for M1 should result in a value of 3.4 kg.
  • #1
Staerke
12
0

Homework Statement



M1 and M2 are two masses connected as shown. The pulley is light (massless) and frictionless.
Find the mass M1, given that M2 (3.5 kg) accelerates downwards at 2.77 m/s2, θ is 25o, and μk is 0.37.

Homework Equations



F = ma
F = mg
Ff = μkN

The Attempt at a Solution



Here's what I've done:
Calculated the force acting on m2 (hanging block)
Fg=3.5*9.81
Fg=34.335
Fma=3.5 * 2.77
Fma=9.695
T=Fg - Fma
T=24.64

Now on to second block
I know these forces are acting on it:
X-direction:
T - Ff - x component of gravity
Y direction
N - y component of gravity

Gravity components:
x = m*g*sin(θ), y = m*g*cos(θ)

I'm only concerned about the forces in the X direction. I know it's accelerating at 2.77 m/s^2

Here's what I put in my calculator
solve(24.64-m*9.81*sin(25)-.37m*9.81*cos(25)=m*2.77,m)
I get 3.4 kg

I have no clue what I'm doing wrong here. This makes sense to me, I really don't understand what I'm missing. Any advice would be greatly appreciated.
The only thing I can think of is if N =/= m*9.81*cos(25) which would throw me off. But I can't figure out why it wouldn't be.
 

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  • #2
Hi Staerke! :smile:
Staerke said:
Here's what I've done:
Calculated the force acting on m2 (hanging block)
Fg=3.5*9.81
Fg=34.335
Fma=3.5 * 2.77
Fma=9.695
T=Fg - Fma
T=24.64

Here's what I put in my calculator
solve(24.64-m*9.81*sin(25)-.37m*9.81*cos(25)=m*2.77,m)
I get 3.4 kg

I have no clue what I'm doing wrong here. This makes sense to me, I really don't understand what I'm missing. Any advice would be greatly appreciated.

(I haven't checked the arithmetic, but …) the method looks ok to me. :confused:

Why do you think it's wrong?
The only thing I can think of is if N =/= m*9.81*cos(25) which would throw me off. But I can't figure out why it wouldn't be.

If we do F = ma in your y direction, a = 0, so your N = mgcosθ is correct.
 
  • #3
tiny-tim said:
Hi Staerke! :smile:


(I haven't checked the arithmetic, but …) the method looks ok to me. :confused:

Why do you think it's wrong?


If we do F = ma in your y direction, a = 0, so your N = mgcosθ is correct.

CAPA (the online program my school uses for physics) says it's wrong... Starting to question whether I believe it.
 
  • #4
Staerke said:
Here's what I put in my calculator
solve(24.64-m*9.81*sin(25)-.37m*9.81*cos(25)=m*2.77,m)
I get 3.4 kg

I've checked the arithmetic now, and i don't get 3.4 :redface:

(I used 3.5(g - 2.77)/{g(sin25° + 0.37cos25°) + 2.77} … is that the same?)

(standard check … you did use ° and not rad?)
 
  • #5


I would recommend checking your calculations and making sure you are using the correct equations. It seems like you are using the correct approach, but your final answer may be incorrect due to a miscalculation or using the wrong equation. Additionally, it would be helpful to draw a free body diagram for each mass to clearly see all the forces acting on them. This can help you identify any mistakes in your calculations or equations. You can also try plugging in your values and equations into a physics calculator or software to double check your answer. Remember to always check your work and make sure it makes sense in the context of the problem.
 

1. What is a block on a pulley on an incline plane?

A block on a pulley on an incline plane is a simple machine system consisting of a block or object, a pulley, and an inclined plane. The block is attached to a rope or cable that is threaded through a pulley and connected to a weight. The incline plane is a flat surface that is tilted at an angle, which allows the block to move up or down when the weight is pulled or lowered.

2. How does a block on a pulley on an incline plane work?

The block on a pulley on an incline plane system works by utilizing the principles of mechanical advantage and gravity. The inclined plane reduces the amount of force needed to lift the weight, while the pulley redirects the force and allows for a longer distance to be covered with less effort. This system is able to lift heavier weights with less force and can also change the direction of the force applied.

3. What is the formula for calculating the mechanical advantage of a block on a pulley on an incline plane?

The formula for calculating the mechanical advantage of a block on a pulley on an incline plane is MA = length of incline plane / height of incline plane. This means that the longer the incline plane is compared to its height, the greater the mechanical advantage will be. For example, an incline plane with a length of 10 meters and a height of 5 meters will have a mechanical advantage of 2 (10/5 = 2).

4. What are some real-life applications of blocks on pulley on incline plane?

Blocks on pulley on incline plane systems have many practical applications in everyday life. Some examples include elevator systems, where a pulley and inclined plane are used to lift and lower the elevator car, and ramps used for loading and unloading heavy objects onto trucks or into buildings. These systems are also commonly used in construction, such as in cranes and hoists.

5. What are the advantages of using a block on a pulley on an incline plane system?

There are several advantages to using a block on a pulley on an incline plane system. First, it allows for the movement of heavy objects with less effort and force. It also allows for the direction of force to be changed, making it easier to lift objects to higher heights. Additionally, this system is relatively simple and inexpensive to construct and can be used in various applications.

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