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# Find The Accelerations A₁, A₂, A, Of The Three Blocks Shown In Figure, If A 1 Horizontal Force Of 10 N Is Applied On (A) 2 Kg Block, (B) 3 Kg Block, (C) 7 Kg Block. Take G = 10 M/S².

20 Answers

Find the accelerations a₁, a₂, a, of the three blocks shown in figure, if a 1 horizontal force of 10 N is applied on (a) 2 kg block, (b) 3 kg block, (c) 7 kg block. Take g = 10 m/s².

# Answers

Best Answer

No best answer has yet been selected by Gauravohcphcphcp. Once a best answer has been selected, it will be shown here.

For more on marking an answer as the "Best Answer", please visit our FAQ.We had to learn alot of tech formulas like friction is F = μR (horizontaly) and F =mg.... stick in your mind forever/

R is the verticle weights in your Q so m x g. and your μs are shown in your diagram if its the same as dizmos, but use whatever yours shows.

I wont spoil the rest for you (Ok hands up I can't do it all them year's later, never was top student, but you'll have some examples in your books???)

FN1 FN2

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F = 10 N | | Fr | |

---------->| 7kg | Fr = 0.3 x 3 x 10 -> Fr = 9 N but also Fr ` = m x a therefore 9N = 3kg x a

So max acceleration before 3kg block slips is 3 m/s^2

This translates to a Max Force (before slippage) of 10kg (7kg + 3kg) x 3m/s^2 = 30 Newtons.

So with 10 N applied we know 3kg block will not slip.

F + Fr = ma -> (10 + 9)/7 = a --- BODMAS --- 19/7 acceleration with 10N applied ---- Answer 2.7 m/s^2

General formula for maximum force F on frictionless surface;

F = µ m2 g (1 + m1/m2) Newtons

Where static friction µ = 0.3, m2 = 3kg , m1 = 7kg and g = 10m/s²

Thus 0.3 x 3kg x 10m/s² (1 + 7kg/3kg) = 30 N

This leads to maximum acceleration before slippage occurs of 3 m/s² as you have correctly calculated.

Where you have gone wrong is to assume the Frictional Force (horizontal vector) is acting Left to Right. The problem is rather intuitive and upon reassessing you’ll see the horizontal vector must be acting from Left to Right. NB: It will alternate in direction as you go up the stack.

One further amendment that is required, not to assume your Frictional Force (correct in the first part) is of the same amplitude for the second part of the question.

Hope this helps.

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