A block of mass `m` is pressed against a wall which is moving with an acceleration a as shown in figure If the block is pulled at an angle `theta` with downward vertical by a force `F` and it does not slide relative to the wall . Find the
a. friction force between the block and wall
b.inimum magnitude of `F` so as to cayse a relative sliding between the block and wall

Answer

a.`f = sqrt(F^(2) + m^(2) g^(2) + 2f mg cos theta)`
b.`F = -mg cos theta + sqrt(m^(2) g^(2)cos^(2) theta - m^(2) (g^(2) - mu^(2) a^(2)))`

Explanation

Adopting ground frame method we have mentioned normal reaction N applied force F gravity `mg` and friction f (arbitearity assumed) in the `FBD` (fig(a and b)].

Force equation :Appling Newton 2nd law on the block , we have `f + sum vec F_(||) = m vec a_(||)` where `vec a _(||) = 0` because the block does not side vertically. This given `vec f + sum vec F_(||) = 0` where `sum vec F_(||) = mvec(g) + vec F` This yieds `vec f + vec F + m vec g = 0`. Hence we have `vec f + - (vec F +m vec g)` as the angle between `vecF` and `m vec g` is `theta`we have
`f = sqrt(F^(2) + m^(2) g^(2) + 2F mg cos theta)`
where` vec F` is direction opposite to the resulting of `vec F` and `m vec g`
b. Force equation : Appling law of motion perpendicular to the plane we have `N = ma`
Law of static friction : The gives as the magnitude of limiting friction `f_(max) = mu_(s) N = mu_(s) ma`
where `f_(s_(max)) = sqrt(F^(2) + m^(2) g^(2) + 2F mg cos theta) `
Then we have `sqrt(F^(2) + m^(2) g^(2) + 2F mg cos theta = mu_(s) ma)`
`F = -mg cos theta + sqrt(m^(2) g^(2)cos^(2) theta - m^(2) (g^(2) - mu^(2) a^(2))`