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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.
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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.
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Does it (the diagram) look anything like this? https:/
done this stuff at night school in the 80s
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???)
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???)
I will attempt to work out 2 BLOCK STACK which hopefully will be of help to Gauravohphcphcp.
FN1 FN2
| |
| |
| |
---------------------- ------------
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
FN1 FN2
| |
| |
| |
---------------------- ------------
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
Kuiper, The first part is correct and complies with the general formula for a 2 block stack where the bottom block (7kg) is on a frictionless surface i.e., there is no static friction hence µ = 0.
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.
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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