I know the answer but I do not understand this part. Please answer the following

Question

I know the answer but I do not understand this part. Please answer the following question. Also, did I do it right?

For 2M - Why is it F cos 30 - T = 2ma?

If F cos 30 is pointed to the left should it be - F cos 30.

If T is pointed to the right should it be - T

9. If F = 40 N and M = 2.0 kg, what is the magnitude of the acceleration of the suspended object? All surfaces are frictionless For 2M-Why is it F cos 30-T = 2ma? If F cos 30 is pointed to the left should it be - F cos 30. If T is pointed to the right should it be- T For 2M For M T-mg = ma F cos 30 T+T -mg2ma + ma Foos 30-T+T-mg = a (2m + m) F cos 30-mg (3m) (a) 40 cos 30-(2) (9.8) = (3) (2) (a) 34.64-19.62 (6) (a) 15.04 = (6m) (a) 2.50 = a F sin 30 cos 30 mg mg

Explanation / Answer

Actually in this case the direction of acceleration is taken towards left of 2m block that's why all the forces along left woll taken as positive and FcosQ is towards left thatst why it is taken as positive and T is towards left That's why it is taken as negative.

In general when you write the equation from free body diagram sign of force is totally dependent on direction of acceleration chosen by us.The forces which are in the direction of acceleration are taken as positive and opposite to acceleration are taken as negative.

And finally forces are always taken in left hand side of equation and acceleration with mass multiplication will always taken in right hand side of equation.

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