## Chapter 7

## Q. 7.1

A projectile *S* weighing 50 lb strikes a concrete bunker with a normal velocity of 1288 ft/sec (878 mph). The projectile imbeds itself in the wall and comes to rest in 10^{-2} sec. What is the average force exerted on the wall by the projectile during this time?

## Step-by-Step

## Verified Solution

The change in the linear momentum of the projectile is

\Delta \mathbf{p}=-\frac{50}{32.2}(1288) \mathbf{n}=-2000 \mathbf{n} \text { slug } \cdot \mathrm{ft} / \mathrm{sec}, (7.5a)

in which **n** is the unit normal vector directed into the wall. The average force \mathbf{F}_S^* acting on the projectile in the time \Delta t=10^{-2} sec is given by the last ratio in (7.4), and with (7.5a) we thus obtain

\mathbf{F}^* \equiv \frac{1}{\Delta t} \int_{t_0}^t \mathbf{F}(t) d t=\frac{\mathscr{G}\left(t ; t_0\right)}{\Delta t}=\frac{\Delta \mathbf{p}}{\Delta t} (7.4)

\mathbf{F}_S^*=-\frac{2000}{10^{-2}} \mathbf{n}=-2 \times 10^5 \mathbf{n} \mathrm{lb}=-100 \mathbf{n} \text { tons }. (7.5b)

This estimates the total force exerted on the projectile by the concrete wall and gravity. Of course, the weight of the projectile compared with the total impulsive force (7.5b) is negligible, and hence the average force exerted on the wall by the projectile may be estimated by the equal and oppositely directed force \mathbf{F}_W^* = lOO**n** tons. If the action time increment is smaller, the average force acting on the projectile or the penetration force acting on the wall grows larger.