Blocks on pulley on incline plane

  • Thread starter Staerke
  • Start date
  • #1
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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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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Hi Staerke! :smile:
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
12
0
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
tiny-tim
Science Advisor
Homework Helper
25,832
251
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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