Solving a Homework Problem with Mass, String, and Friction

AI Thread Summary
Two blocks of mass m_A and m_B are connected by an ideal string, with block A descending at a constant speed and no friction involved. The discussion centers on determining the correct relationship between the masses based on the forces acting on them, particularly using Newton's laws. The initial reasoning suggests that since the speed is constant, the forces must balance, leading to the conclusion that m_A (1/2) = m_B (sqrt(3)/2). After further analysis, it is confirmed that the first answer is correct, as 1/sqrt(3) falls between 0.25 and 0.6. The solution emphasizes the importance of breaking down gravitational components to arrive at the correct answer.
SqueeSpleen
Messages
138
Reaction score
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]
upload_2017-5-19_0-33-15.png

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]
 

Attachments

  • upload_2017-5-19_0-39-4.png
    upload_2017-5-19_0-39-4.png
    9.4 KB · Views: 419
Physics news on Phys.org
Working it out very quickly, and thus prone to error, I think the first answer is correct.
 
SqueeSpleen said:
it would fall in item 1 as 1/sqrt(3) is between 0.25 and 0.6
Right?
Yes.
 
Absolutely correct! Way of solving by simply taking their Sine components of 'g' and solve it.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top