Question

A crate of books is to be put on a truck with the help of some planks sloping up at 36◦ . The mass of the crate is 80 kg, and the coefficient of sliding friction between it and the planks is 0.8. You and your friends push horizontally with a force F~ . The acceleration of gravity is 9.81 m/s 2 .

Answer 1

F= 2.86KN

Step-by-step explanation:

The attached diagram shows the motion of crate on a single line The forces acting on it have been resolved as shown in the attached figure.The kinetic friction opposes the motion.Now crate is moving with constant speed so the acceleration is constant.

Applying Newton's second law to determine the force F and also substituting the values for
`F_(k)` to eliminate normal force

`$\sum_`F = ma

Now
`a_(x)` and
`a_(y)` components are zero

Horizontal and vertical forces are Force on crate, weight of crate and kinetic friction

`$\sum F_(x) = Fcos\theta-f_(k)-mgsin\theta=0`

`$\sum F_(y) = F_(n)- Fsin\theta-mgcos\theta=0`

therefore
`f_(k) = \mu_(k) F_(n)`

`Fcos\theta- \mu_(k) F_(n)-mgsin\theta=0` >>> (1)

`F_(n)=Fsin\theta+mgcos\theta`

By putting the above equation in equation 1 we get F

`F=(mg(sin\theta+\mu_(k)cos\theta) )/(cos\theta-\mu_(k)sin\theta )`

By putting values we get

F =
`((80kg)(9.81(m)/(s^(2) ))(sin36+(0.8)cos36) )/(cos36-(0.8)sin36)`

F= 2860.8 N

F= 2.86KN