I'm stuck Not sure where to go. Help

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Homework Help Overview

The problem involves a system of two masses connected by a cable over a pulley, with one mass resting on a horizontal surface and the other hanging. The context includes static and kinetic friction coefficients, and the goal is to determine the acceleration of the system when it is set in motion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply Newton's second law to both masses, using equations that incorporate tension and friction. Some participants question the initial conditions regarding friction, while others provide feedback on the calculations and units.

Discussion Status

The discussion includes attempts to clarify the setup of the problem, particularly regarding the presence of friction. Some guidance has been offered regarding the correctness of the calculations, but there is no explicit consensus on the final answer.

Contextual Notes

There was an initial misunderstanding about whether the surface was frictionless, which was later corrected. Participants are also discussing the importance of units and significant figures in the calculations.

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Homework Statement


An object with a mass of 10 kg (this mass is m1) rests on a horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass of 4.0 kg (this mass is m2). The coefficient
of static friction between m1 and the horizontal surface
is 0.50, and the coefficient of kinetic friction is 0.30. If the system is set in motion with m2 moving
downward, what will be the acceleration of the system?

Homework Equations


m1 = mass one
m2 = mass two
T = tension
μ = coefficient of friction
g = gravity
a = acceleration
R = normal reaction

I got this helpful hint at another site, but I still can't figure it out...

For m2:
m2 g - T = m2 a ...(1)

Resolving horizontally and vertically for m1:
T - μR = m1 a ...(2)
R = m1 g ...(3)

Substitute for R from (3) in (2):
T - μm1 g = m1 a ...(4)

Add (1) & (4):
(m2 - μm1)g = (m1 + m2)a
a = (m2 - μm1)g / (m1 + m2)
= (4 - 0.3 * 10)g / (10 + 4)
= g / 14
= 0.701 m/s.



The Attempt at a Solution



And this is where I run into the problem. I'm not sure what to do, even with the helpful hint provided on another site.
 
Last edited:
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You initially stated that m1 rests on a frictionless surface, then you give coefficients of static and kinetic friction for m1 and the surface. How is this so?
 
Oops, that was my mistake. I've edited the problem - it isn't frictionless. I'm very sorry about that...
 
Don't sweat it, I just wasn't sure what the problem was trying to accomplish. I'm not sure what you're having trouble on, you've already arrived at the correct answer. The only problem is your units and sig figs. It should be .70 m/s^2.
 
Oh, okay. Thanks a for the assistance.
 
\ W_2 - F_f = (m_1+ m_2) \cdot a eq (1)

\ N = m_1 \cdot g eq (2)

\ F_f = \mu \cdot N eq (3)

\ F_f = \mu \cdot m_1 \cdot g eq (4) from (2) into (3)

\ W_2 = m_2 \cdot g eq (5)

\ m_2 \cdot g - \mu \cdot m_1 \cdot g = (m_1+ m_2) \cdot a

\frac{\((m2 - \mu \cdot m1)} {\((m2 +m1)} g = 0.701
 

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