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- Thread starter sahilmm15
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sahilmm15

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I have drawn all the Free body diagrams finding difficulty in equating them

- #4

sahilmm15

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The diagonal arrows are forces from normal reaction. I didn't wrote it. Sorry for any inconveniences.I have drawn all the Free body diagrams finding difficulty in equating them

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- #6

Lnewqban

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How much acceleration the smaller block would reach?

What forces would induce that acceleration and how strong they would be?

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VEReade

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2)The other two don't look right. For example, the triangular mass M' is accelerating with same acceleration as M to the right. There is also a reaction force on M' due to mass m to consider.

3)To solve this problem, you need to think about what it means acceleration-wise for the m not to slipping over M'

4)Note, in these kind of questions, although you'll always be applying 'F=ma", there will often be more than one way of getting the right answer!

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Without some stated notation, I'm not quite sure what your diagrams mean. You show some actual forces, like mg, but also resultants, like ma, and omit to label the normal forces. It would be clearer had you got as far as writing equations.The diagonal arrows are forces from normal reaction. I didn't wrote it. Sorry for any inconveniences.

Please label all actual forces with variable names and either remove resultant forces or distinguish them by using a double arrowhead.

Or just post the equations you get for force balance,

- #9

Chestermiller

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- #10

sahilmm15

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Well , after some time (30 minutes) I was able to solve the problem. I completely left the problem for 1 hour so that my brain can work on it subconsciously(as suggested by one of my mentors in other thread). It came to my surprise that I solved it easily. Shall I provide you my work?

Last edited:

- #11

sahilmm15

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2)The other two don't look right. For example, the triangular mass M' is accelerating with same acceleration as M to the right. There is also a reaction force on M' due to mass m to consider.

3)To solve this problem, you need to think about what it means acceleration-wise for the m not to slipping over M'

4)Note, in these kind of questions, although you'll always be applying 'F=ma", there will often be more than one way of getting the right answer!

- #12

sahilmm15

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- #13

sahilmm15

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Also for triangular block I didn't draw the fbd but directly wrote the equation.

- #14

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The final step is wrong.##M=M\tan(\theta)+(M'+m)\tan(\theta)##

##M=\frac{M'+m}{\cot(\theta)}##

- #15

sahilmm15

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Thanks for your reply, I appreciate it. It should be cot(x) - ##1## in the denominator. It was an calculation error.The final step is wrong.

- #16

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Yes.Thanks for your reply, I appreciate it. It should be cot(x) - ##1## in the denominator. It was an calculation error.

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