Solving a Homework Problem with Mass, String, and Friction

In summary: If the 'm' on the left side of the equality is neglected, then the problem seems to be impossible to solve.In summary, the conversation discusses two blocks of different masses connected by an ideal string, with one block descending at a constant speed due to the absence of friction. The question is which of the given options is true, with the correct answer being the first option. The reasoning behind this is that since the speed is constant, there is no acceleration, and thus the sum of the forces acting on the system is zero. By decomposing gravity into its components, the solution can be found using trigonometry.
  • #1
SqueeSpleen
141
5

Homework Statement


There's two blocks of mass m_{A} and m_{B} which are linked by an ideal string. The block of mass A descends at constant speed. There's no friction. If the pulley is ideal, which one of the following is true[/B]
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Homework Equations


Newton laws and trigonometry I think.[/B]

The Attempt at a Solution


I think that the correct answer is the second one, but the first one is marked as correct so I wanted a second opinion.
My reasoning is the following: If the speed is constant it means there is no acceleration, so the sum of the forces is 0. Then I decomposed gravity in the component that's compensated by the normal force done by the triangle and it's perpendicular component. So I arrived to
m_{A} (1/2) = m_{B} sqrt(3)/2
Now that I check, it isn't item 2. it would fall in item 1 as 1/sqrt(3) is between 0.25 and 0.6
Right?

sorry for the typos, I broke my keyboard yesterday (I won't drink coffee while on computer again) and I'm with a rubber one until a new one arrives, at least it's better thanthe one that windows have to use with the mouse.

[/B]
 

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  • #2
Working it out very quickly, and thus prone to error, I think the first answer is correct.
 
  • #3
SqueeSpleen said:
it would fall in item 1 as 1/sqrt(3) is between 0.25 and 0.6
Right?
Yes.
 
  • #4
Absolutely correct! Way of solving by simply taking their Sine components of 'g' and solve it.
 
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