A uniform string of length 10 m and mass 20 kg lies on a smooth friction-less inclined plane. A force of 200N is applied as shown in the figure.

(a) Find the acceleration of the string.
(b) Find the tension in the string at 2m from end A.

Explanation

Acceleration of slope, `a=(F-mg sin 30)/(m)`
`=(200 -20 xx 10 xx 1//2)/(20)=5ms^(-2)`
Mass of slope AB
`=AB=M_(AB)=((20)/(10)) xx 2 = 4 kg`

From FBD of AB
`T-m_(AB)sin 30^(@)=m_(AB)a`
`implies T-4 xx 10 xx 1/2 = 4 xx 5 implies T=40N`.