Frictional and Tensional Forces on an Inclined Plane

AI Thread Summary
The discussion revolves around understanding the dynamics of two blocks sliding down a 10-degree incline, focusing on the calculations for acceleration and tension. The user initially struggles with the signs in their calculations, receiving a negative acceleration, which leads to confusion about the motion of the blocks. Clarification is provided that the term "magnitude" refers to the absolute value, allowing for a positive interpretation of the acceleration. It is emphasized that the blocks are indeed in motion, and the negative sign indicates deceleration rather than a lack of movement. The tension in the string is confirmed to be correct, reinforcing the understanding of the system's behavior.
parabolic
Messages
6
Reaction score
0
Let me start by stating that this is not a homework question. If I just wanted the answer, it is in the back of my book. What I do need is some understanding of how the answer was reached.

Problem:
Two blocks attached by a string slide down a 10degree incline. Block 1 has a mass m1 = 0.80 kg and block 2 has mass m2 = 0.25 kg. In addition, the kinetic coefficients of friction between the blocks and the incline are 0.30 for block 1 and 0.20 for block 2. Find (a) the magnitude of the acceleration of the blocks, and (b) the tension in the string.

Answers in the back of the book:
(a) 0.96 m/s2
(b) 0.18 N

http://www.mravery.com/images/Physics.jpg


My attempt at a solution:


After drawing free body diagrams for both blocks I have come up with the following:

Block one calculations
Weight force in x direction (Wx)= sin(10) x (.8) x 9.81

Fnety = may
(Normal Force) - cos(10)x (.8) x 9.81 = 0
N = cos(10)x (.8) x 9.81

Friction = .3 x (cos(10)x (.8) x 9.81)

Fnetx = Wx + Tension - FrictionAfter I calculate these, I calculate the same for block 2 ( I will spare you the redundant reading), and obviously the tensional force is opposite. Once i do that I get stuck.

I have been racking my brain for a day and a half. Any help would be greatly appreciated!

Thanks
 
Last edited by a moderator:
Physics news on Phys.org
You have it correct. Now just set Fnetx on block 1 = m1a, and fnetx on block 2 =m2a, and solve for T and a , using the 2 equations to solve for the 2 unknowns.
 
ok so when i do that I get
(a) -.96 m/s2

why do i get a negative? Does the word "magnitude" mean I disregard the direction (sign)?

for (b) i get the correct answer using the negative
 
Oh, I get the same result, but we shouldn't be getting a negative acceelration. Something has gone afoul in this problem statement or book answer. I believe that neither block ever moves, and that there is no tension in the string, but I have to leave now, and my mind is fuzzy, so let's ponder. Or maybe one of the night owls will help out.
 
I also thought that the frictional force was greater than the Weightx on both blocks. Meaning that they wouldn't move. Weird. I have decided to burn my physics book and change my career path from Mechanical Engineering to Arsonist. Thanks for your help.
 
parabolic said:
I also thought that the frictional force was greater than the Weightx on both blocks. Meaning that they wouldn't move. Weird. I have decided to burn my physics book and change my career path from Mechanical Engineering to Arsonist. Thanks for your help.
Good morning. My bad. The problem clearly states that
(1) the blocks are sliding down the plane, and
(2) it asks for the magnitude of the acceleration.
If the blocks were just placed on the plane, then released, nothing would happen; they'd remain at rest and there would be no tension in the string. But in this problem, they have been set in motion down the plane by some applied force, then that force is released. The blocks then slow down , moving down the plane, and ultimately come to rest. The acceleration is .96 UP the plane (hence the negative sign, the blocks are decelerating), but its magnitude is .96. The tension value is also correct. If the tension value came out negative, then each block would accelerate , or remain at rest, independent of each other (you can't have negative tension). Such is not the case here. Don't change your career yet!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top