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Soloha48 [4]
3 years ago
12

ASAP please!! The force vectors on an aircraft are as shown. Find the net (resultant force).

Physics
1 answer:
ioda3 years ago
6 0

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

W_D=3600 N\cdot\sin 27^\circ=1634N

acting in the direction of the Drag, and the plane-perpendicular component is:

W_L=-3600N\cdot\cos 27^\circ=-3208N

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N

The magnitude of the  net force is:

|F|=\sqrt{5366^2+829^2}N = 5430N


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Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

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From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

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              t__{A}} = \frac{l}{v__{A}}}

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=>           t__{A}} =  50 \  s

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             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

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            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

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