What is the force acting on an object inside a spaceship? A $7.5$ kg instrument is hanging by a vertical wire inside a spaceship that is blasting off from rest at the earth's surface. This spaceship reaches an altitude of $208 $m in $20$ s with constant acceleration.Find the force that the wire exerts on the instrument.
I have a lot of confusion here.The approach I take to solve the problem seems to be the wrong one according to the solutions I have seen on other forums (yahoo for instance)
My thought
The acceleration of the spaceship is given by $\vec{x}=\frac{1}{2}\vec{a} \cdot t^2$, substituting the given values I find $\vec{a}=1.04 \text{ m}/\text{s}^2$ which is positive as I am defining upward to be positive.
So I have that the instrument hanging from a wire is being accelerated at $1.04 \text{ m}/\text{s}^2$. But there is also a opposite acceleration given by $\vec{g}=-9.8 \text{ m}/\text{s}^2$.
Thus I have that the net acceleration of the instrument is given by the vector addition of the corresponding accelerations, i.e. $\vec{a_n}=-8.76 \text{ m}/\text{s}^2$.
So the force acting on the wire is $\vec{F}=m \cdot \vec{a_n}=7.5 \text{ kg} \cdot (-8.76) \text{ m}/\text{s}^2=-65.7\text{ N}$.
As already pointed out,my answer is  apparently wrong because I should have added both accelerations (that means taking both to be positive ,regardless of direction).
The correct solution to this problem is really important for me as I am a self-learner and I don't want to learn false stuff which then I will carry over for the rest of my life .
 A: Consider the forces acting on the particle alone. There is the tension and the weight of the particle. So using Newton's Law $$\underline{F}=m\underline{a}\Rightarrow T-mg=ma$$
Hence $T=mg+ma$
A: Hint always draw a FBD rather than doing mentally. So we have tension acting upwards,gravity acting downwards and ofcourse accn of rocjet upwards so by FBD we have $T=mg+ma$ now we  can use kinematical equations thus $s=ut+\frac{1}{2}at^2$ . Now use words $$blasted off=u=0$$ thus we get $208=200.a$ so $a=1.04$ so $T=7.5\times 9.8+7.5\times 1.04=81.3N$ 
A: You should draw a free body diagram of the object & consider all the forces acting on that as follows  
$\bullet$ Tension force (T): acting upward 
$\bullet$ Gravitational force, $F_g=mg$: acting downward
$\bullet$ Inertial force, $F_I=ma$:  acting downward
Now, balancing all the forces acting on the object, one should get 
$$T=mg+ma$$ $$=m(g+a)$$ $$=7.5(1.04+9.8)$$
$$=\color{red}{81.3\ N}\ (\text{upward})$$
Thus, the tension $T=81.3\ N$ will be the upward force which the wire exerts on the instrument. 
