Answer:
The distance between the helicopter and the man changing at that instant is 41.46 ft/sec.
Explanation:
Given:
Distance between the man and the helipad,
=
ft
Vertical distance between the helipad and the helicopter,
=
ft
Helicopter lift off vertically and is raising at a speed of 45 ft/sec.
That can be written as,
= 45 ft/sec
Now from the diagram (attached with) we can find the hypotenuse of the triangle that is the distance between the man and the helicopter.
Applying Pythagoras theorem:
⇒ ![h^2=b^2+p^2](https://tex.z-dn.net/?f=h%5E2%3Db%5E2%2Bp%5E2)
⇒ ![h=\sqrt{b^2+p^2}](https://tex.z-dn.net/?f=h%3D%5Csqrt%7Bb%5E2%2Bp%5E2%7D)
⇒ ![h=\sqrt{123^2+52^2}](https://tex.z-dn.net/?f=h%3D%5Csqrt%7B123%5E2%2B52%5E2%7D)
⇒
ft
Now to find the changing distance between the helicopter and the man we have to differentiate the Pythagoras theorem as dh/dt can be obtained from there.
And we know that the distance b is constant so db/dt=0.
Using chain rule and power rule of differentiation.
⇒
⇒ ![2h\frac{dh}{dt} =2b\frac{db}{dt} + 2p\frac{dp}{dt}](https://tex.z-dn.net/?f=2h%5Cfrac%7Bdh%7D%7Bdt%7D%20%3D2b%5Cfrac%7Bdb%7D%7Bdt%7D%20%2B%202p%5Cfrac%7Bdp%7D%7Bdt%7D)
⇒ ![2h\frac{dh}{dt} =0+ 2p\frac{dp}{dt}](https://tex.z-dn.net/?f=2h%5Cfrac%7Bdh%7D%7Bdt%7D%20%3D0%2B%202p%5Cfrac%7Bdp%7D%7Bdt%7D)
⇒
<em>....eliminating the common 2 </em>
⇒
⇒ Plugging the values.
⇒ ![\frac{dh}{dt} = \frac{123\times 45 }{133.5}](https://tex.z-dn.net/?f=%5Cfrac%7Bdh%7D%7Bdt%7D%20%3D%20%5Cfrac%7B123%5Ctimes%2045%20%7D%7B133.5%7D)
⇒
ft/sec
So,
41.46 ft/sec is the distance between the helicopter and the man changing at that instant.
Answer:
a) 20.29N
b) 19.43N
c) 15N
Explanation:
To find the magnitude of the resultant vectors you first calculate the components of the vector for the angle in between them, next, you sum the x and y component, and finally, you calculate the magnitude.
In all these calculations you can asume that one of the vectors coincides with the x-axis.
a)
![F_R=(9cos(30\°)+12)\hat{i}+(9sin(30\°))\hat{j}\\\\F_R=(19.79N)\hat{i}+(4.5N)\hat{j}\\\\|F_R|=\sqrt{(19.79N)^2+(4.5N)^2}=20.29N](https://tex.z-dn.net/?f=F_R%3D%289cos%2830%5C%C2%B0%29%2B12%29%5Chat%7Bi%7D%2B%289sin%2830%5C%C2%B0%29%29%5Chat%7Bj%7D%5C%5C%5C%5CF_R%3D%2819.79N%29%5Chat%7Bi%7D%2B%284.5N%29%5Chat%7Bj%7D%5C%5C%5C%5C%7CF_R%7C%3D%5Csqrt%7B%2819.79N%29%5E2%2B%284.5N%29%5E2%7D%3D20.29N)
b)
![F_R=(9cos(45\°)+12)\hat{i}+(9sin(45\°))\hat{j}\\\\F_R=(18.36N)\hat{i}+(6.36N)\hat{j}\\\\|F_R|=\sqrt{(18.36N)^2+(6.36N)^2}=19.43N](https://tex.z-dn.net/?f=F_R%3D%289cos%2845%5C%C2%B0%29%2B12%29%5Chat%7Bi%7D%2B%289sin%2845%5C%C2%B0%29%29%5Chat%7Bj%7D%5C%5C%5C%5CF_R%3D%2818.36N%29%5Chat%7Bi%7D%2B%286.36N%29%5Chat%7Bj%7D%5C%5C%5C%5C%7CF_R%7C%3D%5Csqrt%7B%2818.36N%29%5E2%2B%286.36N%29%5E2%7D%3D19.43N)
c)
![F_R=(9cos(90\°)+12)\hat{i}+(9sin(90\°))\hat{j}\\\\F_R=(12N)\hat{i}+(9N)\hat{j}\\\\|F_R|=\sqrt{(12N)^2+(9N)^2}=15N](https://tex.z-dn.net/?f=F_R%3D%289cos%2890%5C%C2%B0%29%2B12%29%5Chat%7Bi%7D%2B%289sin%2890%5C%C2%B0%29%29%5Chat%7Bj%7D%5C%5C%5C%5CF_R%3D%2812N%29%5Chat%7Bi%7D%2B%289N%29%5Chat%7Bj%7D%5C%5C%5C%5C%7CF_R%7C%3D%5Csqrt%7B%2812N%29%5E2%2B%289N%29%5E2%7D%3D15N)