Question

You are lowering two boxes, one on top of the other, down the ramp shown in the figure (Figure 1) by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 11.0 cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.430, and the coefficient of static friction between the two boxes is 0.838.

What is the magnitude f of the friction force on the upper box?

What force T do you need to exert to accomplish this?

Answer 1

Solution

Data

We can be resolve, this problem doing a Free Body Diagram for upper box and lower box, so follows

1) FBC of upper box( mass 1)

In reference at last figure, the boxes moves are in the negative horizontal axe,

Applying the Newton Law for the mass.Considering the aceleration of de boxes cero, so the speed is constant.

I) Friccion Force.

a)    (1)

(2)

b) (3)

(4)

But, the friction force bewteen the upper and lower box is determined

(5)

For the equation (4), We have the norma bewteen the lower box with respect the upper box

(6)

Susbstituying (6) in (5)

(7)

Where

(8)

: The force normal bewteen the lower box with respect upper box

Where y and x are height and length of the incline plane

II) The Tension Force

2) Now, The FBC of the lowee box (mass 2)

The Newton Law for the mass 2

(9)

(10)

(11)

(12)

(13)

Where

: Normal force between the incline plane surface and lower box.

: Normal force between the upper box with respect the lower box

Clearing   of the equation (13) and substituting in the equation (11), we obtain one expression for the force tension.

(14)

(15)

But

Then

(16)

(17)

Finally

(18)

With

(19)