Question

A safe weighing 2000 N is to be lowered at constant speed down skids 4 m long, from a truck 2 m high. (a) If the coefficient of sliding friction between safe and skids is 0.3, (a) will the safe need to be pulled down or held back? (b) How great a force parallel to the skids is needed?

Answer 1

Draw the situation:

Find the sine of the angle $\theta$ (theta) for the right triangle where the height is 2 m and the hypotenuse is 4 m:

$$\text{ sin}\theta=h/l=2/4=0.5$$

(a) Find the acceleration of the safe using Newton's second law:

$$Ox:ma=-f+mg\text{ sin}\theta,\\ Oy: N-mg\text{ cos}\theta=0,\\ f=\mu N.$$

Hence find the normal force, the force of friction and, thus, the acceleration:

$$N=mg\text{ cos}\theta,\\ f=\mu mg\text{ cos}\theta,\\ a=g\text{ sin}\theta-\frac{f}{m}=g(\text{sin}\theta-\mu\text{ cos}\theta)=2.35\text{ m/s}^2.$$

(b) This answer means that without any external control, the safe will slide down increasing its velocity for 2.35 m/s every second. Therefore, an additional external force F is required to make the acceleration 0. We can use the equation for Ox to find this force. Of course, as we see in the image, this force must act upward along the incline (in the direction of f). But this is a very long way. It is much easier to find this force from the condition that it must stop a safe with mass m=(2000/9.8) kg from sliding with acceleration of 2.35 m/s²:

$$F=ma=\frac{W}{g}a=\frac{2000}{9.8}2.35=480\text{ N}.$$