# Homework Help: A simple question about Force

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1. Jan 18, 2017

### Arman777

1. The problem statement, all variables and given/known data
The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air.In one version of the act,Emanuel Zacchini was shot over three Ferris wheels to land in a net at same height as the open end of the cannon and at a range of 69 m.He was propelled inside the barrel for 5.2 m and launched at an angle of 530 ( degree).If his mass was 85 kg and he underwent cosntant acceleration inside the barrel,what was the magnitude of the force propelling him ? ( Hint:Treat the launch as though it were along aa ramp at 530 ( degree).Neglect air drag )

2. Relevant equations
$$\vec F_t=m \vec a$$ $$m=mass$$ $$\vec F_t =Total \ Force$$
$$\vec w_=m \vec g$$ $$g=9.8 m/s^2$$
$$v-v_0=at$$ $$v_0=inital \ velocity$$

Projectile motion equations (In this question they will be in different forms I guess and I dont know how to find them thats my problem)

3. The attempt at a solution

I upload a picture.After that solution I am stucked.Cause theres acceleration and also velocity of the object and I have to apply them to projectile motion equations.

Here my try
For x axis ;

$$69m=v_0cos(53^0) t+\frac 1 2 acos(53^0)t^2$$
Simply I couldnt solve.

For y axis;
Here things are more complicated
Simply I couldnt solve.

Also another referance frame would be easier to solve but I couldnt put the forces in there something's doesnt make sense.I would be happy If somebody help me
Thanks

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2. Jan 18, 2017

### phinds

3. Jan 18, 2017

### Arman777

Then this is the best thing I can do

Here F is The force that cannonball applies, I showed my referance frame, m is mass of object, a is the acceleration of object inside the cannonball.

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4. Jan 18, 2017

### Arman777

In this referance frame I can write the components of force.But is it a better then my pre-referance frame...
Components are ;
w=mg (-y)
N=Ncos37 (x) + Nsin37(y)
F=Fcos53 (-x) + Fsin53(y)

so Ftotal=ma cos53 (-x) + ma sin53 (y)

so ;
-ma cos53=Ncos37-Fcos53
and
ma sin53 =-mg+Nsin37+Fsin53

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5. Jan 18, 2017

### haruspex

Deal with the trajectory first. Find the KE with which he must emerge from the barrel.

6. Jan 19, 2017

### Arman777

I can use only Newtons's Law.
And I just dont wanna solve the question and leave there.I want to know where am I doing wrong.
Thanks

7. Jan 19, 2017

### haruspex

You will have to deal with the trajectory at some point, and that also uses Newton's laws.
It will be much easier if you use a coordinate parallel to the ramp. Try writing the F=ma equation for that.
You won't need to work with a second coordinate.

8. Jan 19, 2017

### Arman777

I didnt want to use energy cause the book hasnt arrived there yet.So this question has a solution without using any energy transformation.(If that was your point)
So theres two equations.(Theres a pic in my #3 post A simple question about Force )
$-ma = - F + mgsin(53^0)$ (Eq-1)
$F:\text{The force that we wanted to know}$
$a:\text{acceleration of object}$
$N=mgcos(53^0)$ (Eq-2)

9. Jan 19, 2017

### haruspex

Ok, that relates F to a. Next, you need to relate a to the launch speed. What equations do you know that would do that?

10. Jan 20, 2017

### Arman777

$v^2-(v_0)^2=2a(x-x_0)$ So here $v_0=0$ and $x_0=0$
so If we use this equation to in our referance frame we willl get

$v^2-0=2(-a)((-x)-0)$ which equals to;
$v^2=2ax$

11. Jan 20, 2017

### Arman777

Ok I solved the problem.Just I made a simple false assumption.And then I couldnt solve cause of that.Now I solved thanks again