Question

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is flying straight down, its speed is 55 m/s and it is speeding up at a rate of 12 m/s per second.

a. what is the magnitude of the net force on the plane you can neglect air resistance.

b. what angle does the net force make with the horizontal let an angle above horizontal be positive and an angle below horizontal be negative

Answer 1

a) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force  $F=m \frac{v^2}{r}$, directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$, so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$, we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$
 

Answer 2

a) R = $\rm 2.23\times 10^6 \; N$

b) $\theta = -26.8^\circ$

Given :

Mass = 85000 Kg

Diameter = 260 m

Speed = 55 m/sec

Solution :

a) There is centripetal force acting on plane is,

$\rm F_c = \dfrac {mv^2}{r}$

$\rm F_c = \dfrac{85000\times 55^2}{130} = 1.98\times 10^6 \; N$

Now weight of the plane is

W = mg

$\rm W = 85000 \times 9.81 = 8.34 \times 10^5 \; N$

Now from Newton's second law,

$\rm F_t = ma$

$\rm F + mg = ma$

$\rm F = m(a-g)$

$\rm F = 85000(12-9.81) = 1.85\times 10^5\; N$

So, the net force acting on the plane will be,

$\rm R=\sqrt{F_c^2 + (W+F)^2} =\sqrt{(1.98\times 10^6)^2 + (8.34\times 10^5 + 1.86\times 10^5)^2}$

$\rm R = 2.23\times 10^6 \; N$

b) The net force make angle with the horizontal is,

$\rm tan\theta = -\dfrac{W+F}{F_c}=-0.5$

$\theta = -26.8^\circ$

For more information, refer the link given below

https://brainly.com/question/11324711?referrer=searchResults