Question

You are lowering two boxes, one on top of the other, down a ramp by pulling on a rope parallel to the surface of the ramp (Fig. E5.33). Both boxes move together at a constant speed of 15.0 cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.444, and the coefficient of static friction between the two boxes is 0.800. (a) What force do you need to exert to accomplish this

Answer 1

The force do you need to exert to accomplish this is 56.98 N.

Step-by-step explanation:

Given that,

Speed = 15.0 cm/s

Coefficient of kinetic friction = 0.444

Coefficient of static friction = 0.800

According to figure,

We need to calculate the angle of ramp

Using formula of angle

`\theta=\tan^(-1)((2.5)/(4.75))`

`\theta=27.75^(\circ)`

Now as blocks are moving with no acceleration they can be considered as a system and N be the normal force which applied by wedge on blocks then perpendicular to slope of wedge,

(a). We need to calculate the force do you need to exert to accomplish this

Using balance equation

`T+\mu N=mg\sin\theta`

Here,

`N = mg\cos\theta`

`\mu` = coefficient of kinetic friction

`T=mg(\sin\theta-\mu\cos\theta)`

Put the value into the formula

`T=80*9.8(\sin27.75-0.444\cos27.75)`

`T=56.98\ N`

Hence, The force do you need to exert to accomplish this is 56.98 N.