Block `A and B` shown in the figure are connected with a bar of negligible weight. `A and B` each has mass `170 kg` , the coefficient of friction between `A` and the plane is `0.2` and that between `B` and the plane is `0.4 (g = 10 ms_(-2))`

What is the total force of friction between the blocks and the plane?

Options

A. `900 N`
B. `700 N`
C. `600 N`
D. `300 N`

Answer

A

Explanation

` tan theta = (8)/(15) :. theta = tan^(-1) ((8)/(15)) = 28^(@)`
`(f_(A))_(max) = 0.2 xx 170 xx 10 xx cos 28^(@)`
`= 300.2 N = 300 N`
`(f_(B))_(max) = 0.4 xx 170 xx 10 xx cos 28^(@)`
`= 600.4 N = 600 N`
Now,
`(m_(A) + m_(B)) g sin theta = (340) (10) sin 28^(@) = 1596 N`
Since this is greater than `(f_(A))_(max) + (f_(B))_(max)`, therefore blocks sides downwards and maximum force of friction will act on both surfaces
`:. (f_("total") = (f_(A))_(max) + (f_(B))_(max)`
`= 900 N`