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We have an inclined pulley with angle of $30° $ a block with mass of $10 kg$ placed upon it and It is attached by a rope over a pulley to a mass of $M_A$ which hangs vertically. If $μ_s=0.4$ and $μ_k=0.2$, Then in which mass(es) of $M_A$ the system will not accelerate?(take $g=10 m/s^2$)

I'm not looking for a complete solution on this. I just want to know how I should handle this problem.

My thinkings: First, I considered the case when there is no friction. I found that if $M_A=5 kg$, then the system will not accelerate.

Now my question is:

1- How should I find the situations where the system wants to accelerate but the static friction won't let the box to move?

Diagram enter image description here

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  • $\begingroup$ Add a diagram to make the question visualizable $\endgroup$
    – Sathish
    Dec 19, 2014 at 8:47
  • $\begingroup$ @Sathish,diagram added $\endgroup$ Dec 19, 2014 at 8:54
  • $\begingroup$ The same way you always solve this kind of problem: choose axes and write down Newton. $\endgroup$
    – Wouter
    Dec 19, 2014 at 9:14

2 Answers 2

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Considering that the $10kg$ block does not accelerate from its initial position (ie the mass does not move), you need not have kinetic friction coefficient $\mu_k$ to solve the problem. Just use the static friction coefficient alone.

Free body diagram: enter image description here

where $R=10 g \cos (30^\circ)$ and $\mu=\mu_s$. For equilibrium (zero acceleration), $$Mg-T=0$$

$$ T-10g\sin (30^\circ)-\mu_s (10g\cos (30^\circ))=0$$

(Force balance equations) Solve for M from the above equations to get the value of mass $M$ required!

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First find the force of gravity on the block, in coordinates relative to the inclined plane. "Down" relative to the plane, $F=ma$ yields $$F=10kg*10\frac{m}{s^2}*cos(\pi/6)=70.7N$$ "Left" relative to the plane, we have $$F=10kg*10\frac{m}{s^2}*sin(\pi/6)=50N$$ Since the tension on the string means the force of gravity on the hanging block is the same as the force pulling "right" on the sliding block, and using right as the positive direction, $$F_{net}=10*M_A-50N$$ So if $M_A = 5kg$ then $F_{net}=0$.

Static friction will oppose forces acting in either direction, and the force of static friction is $F_{static}=\mu_s*F_{normal}$. Using values from the problem, $$F_{static}=0.4*70.7N=28.3N$$ So if $|F_{net}|<28.3N$, both blocks will remain stationary.

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