Which block reaches the wall faster?

AI Thread Summary
The discussion centers on the acceleration of two blocks, A and B, connected by a string over a pulley. Block A accelerates faster in the x-direction than block B due to the relationship between their tensions and the angle of the string. The reasoning concludes that since both blocks start from rest and have the same distance to the wall, block A will reach the wall first because its acceleration is greater. Participants suggest that the argument could be presented more clearly, emphasizing the comparison of accelerations. The consensus is that the reasoning is sound, though some believe it could be simplified.
MathVoider
Messages
2
Reaction score
0
Homework Statement
Two block, A and B, of identical mass are connected together by a massless string which is always under tension. They each are of a distance L from the pulley. The pulley has negligable size compared to the blocks. All kinds of friction can be ignored. Initially the block B is suspended by a string. The question is, when the aforementioned string is cut, which block reaches the wall/pulley before the other? In other terms which block travels the distance "L" the fastest.

Scroll down for the diagram:
Relevant Equations
x
normale.PNG

My Solution​

after the string is cut we have a system well defined by the following free body diagram:

problema.PNG


I argue that the acceleration of the block A in the x direction (##a_A^x##) is:
$$ F_A = T_1 $$
$$ ma_A^x = T_1 $$
$$ a_A^x = \frac{T_1}{m} $$
I argue that the acceleration of the block B in the y direction (##a_B^y##) is:
$$ F_B^y = T_2\cos(\theta)-F_g $$
$$ ma_B^y = T_2\cos(\theta)-mg $$
$$ a_B^y = \frac{T_2\cos(\theta)}{m} -g $$
I argue that the acceleration of the block B in the x direction (##a_B^x##) is:
$$ F_B^x = -T_2\sin(\theta) $$
$$ ma_B^x = -T_2\sin(\theta) $$
$$ a_B^x = -\frac{T_2\sin(\theta)}{m} $$

Now that we've got the equation making out of the way i can explain my resoning:
I state that ##T_1## and ##T_2## are egual because when the block B moves along the direction of the string then the block A has to move by the same distance. As such we get:
$$ a_A^x = \frac{T_1}{m}$$ $$a_B^x = -\frac{T_1\sin(\theta)}{m} $$
if we take the absolute value we get:
$$ a_A^x = \frac{T_1}{m}$$ $$a_B^x = \frac{T_1\sin(\theta)}{m} $$
now I argue that ##a_A^x>a_B^x## because ##\sin(\theta)## goes from 0 to 1, since block A and B have the same distance from the pulley\wall, and they both start at rest, and block A has a greater accelleration then block B then block A will reach the pulley\wall faster.

Question​

I want to know if my reasoning is correct or if I have forgotten/misused something.
 
Physics news on Phys.org
MathVoider said:
$$ a_A^x = \frac{T_1}{m}$$ $$a_B^x = -\frac{T_1\sin(\theta)}{m} $$
if we take the absolute value we get:
$$ a_A^x = \frac{T_1}{m}$$ $$a_B^x = \frac{T_1\sin(\theta)}{m} $$
now I argue that ##a_A^x>a_B^x## because ##\sin(\theta)## goes from 0 to 1, since block A and B have the same distance from the pulley\wall, and they both start at rest, and block A has a greater accelleration then block B then block A will reach the pulley\wall faster.
I do not see anything missed, although it seems like more work than is required.

I would have liked to see the point more clearly and directly made that $$|a_A^x| = \frac{T_1}{m} > \frac{T_1 \sin \theta}{m} = |a_B^x|$$
Personally, I would have taken a different approach. Reason first that the horizontal position of the center of mass will accelerate in the direction of the horizontal net force from the pulley. That being a rightward (or briefly zero) net external horizontal force, the center of mass must be to the right of the wall when one of the two blocks first reaches the wall. Clearly, this can only happen if block A reaches the wall first.
 
jbriggs444 said:
I do not see anything missed, although it seems like more work than is required.

I would have liked to see the point more clearly and directly made that $$|a_A^x| = \frac{T_1}{m} > \frac{T_1 \sin \theta}{m} = |a_B^x|$$
Personally, I would have taken a different approach. Reason first that the horizontal position of the center of mass will accelerate in the direction of the horizontal net force from the pulley. That being a rightward (or briefly zero) net external horizontal force, the center of mass must be to the right of the wall when one of the two blocks first reaches the wall. Clearly, this can only happen if block A reaches the wall first.
Thank you
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top