Question

The mass of each car is 600kg, 970kg, 230kg, 810kg, 660kg. The force of friction on each car is 1980N, 3201N, 759N, 2673N, 2178N respectively. What is the magnitude of the upward forces on the 5th car?

Answer 1

If car is on horizontal road then upward normal reaction R on the car is equal to its weight `W= "Mass"(M)xx"acceleration due to gravity"(g)`

So `R=W=Mg`

Again

`"Force of friction"(F)=muxxR=muxxMg`
`"where "mu="coefficient. of friction"`

So for Car 1

`F_1/(M_1g)=1980/(600xx9.80)~~0.34=mu`

So for Car 2

`F_2/(M_2g)=3201/(970xx9.80)~~0.34=mu`

So for Car3

`F_3/(M_3g)=759/(230xx9.80)~~0.34=mu`

So for Car 4

`F_4/(M_4g)=2673/(810xx9.80)~~0.34=mu`

So for Car 5

`F_5/(M_5g)=2178/(660xx9.80)~~0.34=mu`

So for every car Upward reaction `R=Mg`

Hence for 5th car `R_5=M_5g=660xx9.8=6468N`

Answer 2

We know that Normal reaction `N` for a body as shown in the figure below is given by
`N=mgcostheta` ......(1)
Also that force of friction is given as
`f=muxxN=mumgcos theta` .....(2)
where `mu` is coefficient of friction, `m` mass of the object, `theta` is the angle of the inclined on which the object rests and `g` acceleration due to gravity `9.81ms^-2`

Assuming that all cars are located on the same plane and have same coefficient of friction.

For car `1`
`f_1=mum_1gcos theta`
Inserting given values we get
`1980=600xx9.81mucos theta`
`=>mucostheta=0.3364`, rounded to four decimal places

Similarly for car `2`
`3201=970xx9.81mucos theta`
`=>mucostheta=0.3364`, rounded to four decimal places.

Similarly for car `3`
`759=230xx9.81mucos theta`
`=>mucostheta=0.3364`, rounded to four decimal places.

Similarly for car `4`
`2673=810xx9.81mucos theta`
`=>mucostheta=0.3364`, rounded to four decimal places.

We see that value of `mucostheta` in case of all cars is same as expected due to assumption above.

Even though it has been not stated in the problem, it is assumed that all cars are in a state of equilibrium while resting on the plane. As such `theta` is the angle of repose. In such a case we see from the figure that
`f=mgsintheta`
For car `5`
`m_5gsintheta=mum_5gcos theta`
`=>sintheta=mucos theta`
`=>theta=sin^-1(0.3364)`
`=>theta=19.66`
from (1) we have
`N_5=m_5gcos theta`
`N_5=660xx9.81xx0.9417`
`N_5=6097N`