Difficulty with this Problem involving a Pully and Cart

In summary: Are the acceleration magnitudes the same? I don't know how not to mention gravity when that is what is pulling on the mass.Yes, the string always makes a right angle with the cart. And what is the significance of that?Geometrically, they are at 90degree angle? Or atleast the string...But the cart is accelerating. How would you express the acceleration of the cart in terms of the acceleration of the crate?Yes, the string always makes a right angle with the cart. And what is the significance of that?The angle between them changes as the crate moves, but it is always at 90 degrees. How can you write an equation that involves only the accelerations and magnitudes of the
  • #36
i_love_space_and_eng said:
Hm... Okay, they actually do make a lot of sense! My only question is, for the m_1a=m_1g-T_1 equation, what would a be? Because isn't the acceleration of the weight dependant on gravity?
Yes, that's the m1g term, but it is partly countered by the tension in the string.
 
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  • #37
haruspex said:
Yes, that's the m1g term, but it is partly countered by the tension in the string.
Alright... so how would we go on to find the "a" without the T1 and T2?
 
  • #38
i_love_space_and_eng said:
Alright... so how would we go on to find the "a" without the T1 and T2?
You need one more equation, a simple relationship between a and α.
 
  • #39
Okay, so this is what I set up.
I(a/r)= ((m1g-m1a)-m2a)R
Does that look correct? I substituted a/r for α .
 
  • #40
What is I in terms of the mass M and radius R of the pulley?
 
  • #41
Chestermiller said:
What is I in terms of the mass M and radius R of the pulley?
It will be 6.125*10^-5kgm^2
 
  • #42
i_love_space_and_eng said:
It will be 6.125*10^-5kgm^2
Please provide it algebraically.
 
  • #43
I=(2/5)mr2
 
  • #44
i_love_space_and_eng said:
I=(2/5)mr2
So combine the three equations in #32 with those in #39 and #43.
 
  • #45
Okay so I plugged it all in and got this:
(2/5)m(r^2)(a/r)=(((m1g)-(m1a))-(m2a))R
I plug in all the correct numbers but the issue is I still get the wrong answer. Did I write my equation correctly? Should I be getting the right answer and I am probably doing a math error?
 
  • #46
i_love_space_and_eng said:
Okay so I plugged it all in and got this:
(2/5)m(r^2)(a/r)=(((m1g)-(m1a))-(m2a))R
I plug in all the correct numbers but the issue is I still get the wrong answer. Did I write my equation correctly? Should I be getting the right answer and I am probably doing a math error?
Your equation looks right. First simplify it, then post that and your calculation steps.
 
  • #47
i_love_space_and_eng said:
Okay so I plugged it all in and got this:
(2/5)m(r^2)(a/r)=(((m1g)-(m1a))-(m2a))R
I plug in all the correct numbers but the issue is I still get the wrong answer. Did I write my equation correctly? Should I be getting the right answer and I am probably doing a math error?
Those should be capital R's on the left side of the equation, since they are the radius of the pulley. They should also then cancel with the R on the right hand side of the equation. Please provide you algebraic results for a (i.e., in terms of algebraic variables).
 

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