Question

find acceleration and tension in the string for the given diag. angle of inclination is 30 degree andM=2kg. take g=9.8m/s2​

Answer 1

Answer:

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Step-by-step explanation:

Given,

m=2kg

θ=30  

μ=0.5

g=10m/s  2

 The contact force is F  

is the resultant of frictional force and normal force.

From the free body diagram,

Normal force, N=mgcosθ

N=2×10×cos30  

N=17.320N

​ =19.381N

We can say approximately 20 N

Answer 2

Answer:

From FBD of 2 kg block,

2a=2gsin37−T−f

2a=2gsin37−T−0.25(2gcos37)............................(1)

From FBD of 4 kg block,

4a=4gsin37+T−f

′

4a=4gsin37+T−0.75(4gcos37)............................(2)

Eq.(1)+Eq. (2),

6a=6gsin37−3.5gcos37

⇒a=1.3m/s

2

put the value in (1), we will get,

2.6=12−T−4

⇒T=8−2.6=5.4N