1. Oct 3, 2009

### Arliena

Hello, Im new. Im soo despret. I want to become an Architect but I'm dumb. I aws doing fine getting 84% in physics university 11, but now, im taking physics 12 university and I don't get the answers right 65% of the time. Im so despret for help. My parents cannot afford to get me a tutor. At least half of my classmates gets nice marks on the test, but Im soo stupid that i failed the first test. And yes, I did do all my homework. I feel soo stupid its ridiculous. I need serious help.

Question 1) Block A and B are connected by a string passing an essentially frictionless pulley. When the blocks are in motion, Block A experiences a force of kinetic friction of magnitude 5.7N. If Ma= 2.7kg and Mb=3.7kg, calculate the magnitude of

a) The acceleration of the blocks
b) The tension in the string.

2. Oct 3, 2009

### htetaung

hi there
Your question gives inadequate information, whether blocks are on a table or not. But from the available data, it is reasonable to assume that block A is on a table, block B is hanging around the pulley - those are my assumptions.
Under my assumptions, look at the block A first. There is friction(f) and tension(T) in the string. This gives: T - f = Ma*a ___________ 1
Again, look at the block B. There is its weight, Mb*g and tension. This gives:
Mb*g - T = Mb*a __________ 2
Here the two tensions for A and B are equal because the pulley has no friction and acceleration of A and B are equal because they move together - there is no relative velocity between those two.
Solving 1 and 2 gives both tension and acceleration.
Good luck with your studies, Arliena.
Thanks

3. Oct 3, 2009

### Arliena

Hello, Thank you SOO mch for your reply! Im very sorry for my poor discription. And yes, you're right. The block A is on the table. However, I don't understand how i can get the acceleration by using (MB)(g)-T=(MB)(A) because i don't know both Acceleration and the Tension. The texbook says the Answers are:

A) 4.8m/s*2
B) 19N

but I don't know how you got them.

Last edited: Oct 3, 2009
4. Oct 3, 2009

### htetaung

hi there
In the first eq, you know friction and mass of A. For the second eq, you know mass of B and g. This gives you a two simultaneous eq's in two unknowns which I think is easy enough to solve. You have two eq's, not one.

5. Oct 3, 2009

### htetaung

I got them by separately using Newton's second law for each block.
net force = mass times acceleration

6. Oct 3, 2009

### Arliena

Hello, Thanks again for your reply! I tried for A) using (MB)(g)-T=(MB)(A)
(3.7)(9.8)down-(5.7)=(3.7)(a) to (3.7)(9.8)/(3.7) - 5.7 Since 3.7 cancel, Im left with 9.8-5.7 but that doesnt give me 4.8 m/s*2 , it gives me 4.1

7. Oct 3, 2009

### htetaung

hi there again,
T - f = Ma*a __________ 1
Mb*g - T = Mb*a ______ 2
Add 1 and 2, you get: Mb*g - f = (Ma + Mb)a
Plug in everything except a, and you get exactly 4.775
Plug in the value of a in any of the two eq's.

8. Oct 4, 2009

### Arliena

WOW I didn't know that. Seriously... these questions look soo hard but then its simple. THANK YOU VERY VERY MUCH. You don't understand how much you've helped me. Bless you.