Blocks on pulley on incline plane

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Homework Help Overview

The discussion revolves around a physics problem involving two masses connected by a pulley on an inclined plane. The problem requires finding the mass M1, given the acceleration of mass M2, its weight, the angle of incline, and the coefficient of kinetic friction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on both masses, including gravitational forces, tension, and friction. There is an exploration of the equations of motion and the setup of the problem. Some participants express uncertainty about the normal force and its calculation.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning the correctness of their methods. There is acknowledgment of potential errors in arithmetic and a discussion about the validity of the normal force assumption. No consensus has been reached regarding the correct mass value.

Contextual Notes

There is mention of an online program providing feedback that suggests the calculated mass is incorrect, which raises doubts among participants about their approach and calculations.

Staerke
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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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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.
 
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.
 
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?)
 

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