Equation Finding Strategies for Solving Physics Problems

  • Thread starter shin777
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In summary, you need to find 5 equations for each object plus 2 more that you don't know of. You calculated a = 4.1 m/s^2.
  • #1
shin777
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o0ti0m.jpg


so far, I know I have to find 5 equations for each object plus 2 more that I don't know of.
I wrote down everything I know but I just don't understand what to do at this point. please, be more descriptive and help me find equations. So far, I haven't gotten single useful help on the forum.
 
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  • #2
Please keep the units in your equations.

You can related alpha to a via the disk radius.
a and T1 to T4 are 5 unknown constants, (1) to (3) are 3 equations, so you just need two more. You can get those equations from the disk and the ring:
What did you calculate at (4)? T2 would indicate the acceleration of the disk, but then T1 is missing in the equation. In addition, where does the formula for I come from? It is a disk, not a ring.

So far, I haven't gotten single useful help on the forum.
Looking at your previous threads, I get a different impression.
 
  • #3
i got a = 2g(m2 - m1)/(M=2m2+2m1)
a = [2(9.8)x(30-5)]/[5+2(30)+2(5)]
a = 6.53 m/s^s

m1 = 5kg, m2 = 30kg, m3 = 5kg

does it look ok?
 
  • #4
How did you get that?
It would be a very strange coincidence if that would be right, as I don't see how you used the disk and the ring to get that formula. And even then, I think the prefactors are not right.
 
  • #5
I followed my school tutor's lead. but more I look at this problem more I wonder it's wrong. left side of circle should calculated as disc and right side should calculated as ring but on here, it seems he count both as disc and end up having 2 boxes with 1 pulley(disc) instead of 3 boxes with 2 pulley(one disc, one ring). Is it just my imagination or is this just done wrong?
 

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  • #6
That would be the solution if the second pulley would not exist at all.
 
  • #7
that's what i thought. first pulley is disc and second pulley is ring. no matter how many times i look at this, it didn't count second pulley into this calculation.
 
  • #8
Fd = (1/2) Mp r^2 * a / r^2 = (1/2) Mp
Sum of forces = Mg - mg - mg - Fd - Fh = Mg - 2mg - Mp a - (1/2) Mp a = Mg - 2mg - (3/2) Mp a

Sum of forces also = (M + 2m) a

Mg - 2mg - (3/2) Mp a = (M + 2m) a

Mg - 2 mg = (M + 2m + (3/2) Mp) a

a = [(M - 2m) / (M + 2m + 1.5 Mp)] g

a = 0.41 g

a = 4.1 m/s^2

how about this one? this one seems more likely to me.
 
  • #9
There is an "a" missing in the first line.
The way you split those forces is .. unconventional (I would sum all parts with *a and all parts with *g instead of a mixture of both), but the result looks right.
 

1. How do you find the value of "a" in an equation?

To find the value of "a" in an equation, you need to isolate the variable "a" on one side of the equation. This can be done by using inverse operations, such as addition, subtraction, multiplication, or division, to cancel out any other numbers or variables that are attached to "a". Once you have "a" by itself, the remaining numbers will give you the value of "a".

2. Can you give an example of finding the value of "a" in an equation?

Sure, for example, let's take the equation 3a + 12 = 30. First, we need to get rid of the 12 by subtracting it from both sides of the equation. This leaves us with 3a = 18. Then, we divide both sides by 3 to isolate "a". This gives us the answer that a = 6.

3. What do you do if there are multiple "a"s in an equation?

If there are multiple "a"s in an equation, you can still use the same process of isolating the variable "a" on one side of the equation. Just remember to perform the same operation on both sides of the equation to maintain balance. For example, if you have an equation like 2a + 3 = 5a - 12, you can subtract 2a from both sides to get 3 = 3a - 12, and then add 12 to both sides to get 15 = 3a. Finally, divide both sides by 3 to find that a = 5.

4. What if "a" is a fraction or decimal in an equation?

If "a" is a fraction or decimal in an equation, you can still use the same process of isolating the variable. Just remember to perform the same operation on both sides and use the appropriate rules for working with fractions or decimals. For example, if you have an equation like 2a + 1.5 = 3.5, you can subtract 1.5 from both sides to get 2a = 2, and then divide both sides by 2 to find that a = 1.

5. Are there any shortcuts for finding the value of "a" in an equation?

Yes, there are some shortcuts for finding the value of "a" in certain types of equations. For example, if you have an equation in the form of ax + b = c, you can use the formula a = (c-b)/x to find the value of "a" without having to isolate the variable. However, these shortcuts only work for specific types of equations and may not always be applicable. It is always best to use the standard process of isolating the variable to find the value of "a".

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