Question

Help please?

Answer 1

See below.

Explanation:

Considering `theta = 30^@`, `F = 250` and `N=m g` we have

a)
Left `{(sum_X->Fcostheta-mu N=0),(sum_Y->-F sintheta-P+N=0):}`

then `N = F sintheta+P` and the needed force is

`mu N = mu(P+Fsintheta) = 257.25`

Right `{(sum_X->-Fcostheta+mu N=0),(Fsintheta-P+N=0):}`

then `N = P-F sintheta` and the needed force is

`mu N = mu(P-Fsintheta) =22.25`

b)
`(sum_X f) = m a -> F-mu N = m a`

then

`mu = (F-ma)/N = (F-ma)/(m g)= (330 - 35*4.78)/(35*9.81)approx0.474`