What is the force acting on an object inside a spaceship?

Question

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 .

Answer 1

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.

Answer 2

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 $T=mg+ma$

Answer 3

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 $T=7.5\times 9.8+7.5\times 1.04=81.3N$