Tension force in a rope bringing down a mass [closed]

This is the question:

A rope brings a 50 kg object downward at an acceleration of 0.75. What is the tension force of the rope?

I thought that, since the tension and the weight are both "going" downward, the net force would be Weight + Tension = Mass x -Acceleration. The acceleration is negative because it is going downward. But if I do it that way I get the wrong answer. The answer is only correct if net force is Weight - Tension, and if acceleration is positive, neither of which makes sense to me. Can someone explain this?

It could be a silly arithmetic error that I haven't noticed.

closed as off-topic by sammy gerbil, stafusa, John Rennie, Yashas, Jon CusterNov 19 '17 at 16:22

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Because the tension points upwards. Notice that the acceleration you gave is less than $g$, and thus $T$ opposes gravity. Choosing the positive axis up, the correct equation is

$$T-mg=ma$$

with $a<0$ as you stated.

I suspect it is an error your understanding of how the problem the problem is set up.

If there was no rope, the mass would accelerate downward at about $10 m/s^2$.

If the acceleration is smaller than that, the rope must be pulling upward.

This may be enough. If not, force and acceleration are vectors. The direction matters. See this question. Why is the tension between two masses connected by a rope and undergoing a force along the direction of the rope less than that force?

It’s given that the system (rope+block) is accelerating downwards. Acceleration is the keyword here. Attach a frame to the system such that with respect to it the system appears to be at rest, let’s name it S. It is assumed that a person on the ground is in the nearly inertial ground frame of refernce,let’s name it K. Now, with respect to K, S is a non inertial frame and K is an inertial frame. Hence Newton’s Laws of motion don’t work in here(S). So, we have to consider the so called pseudo forces on the system in S. Now in S, taking the rope as the system in consideration, we have the block pulling the rope downwards with force, F=-mg (m is mass, g is acceleration due to gravity, - denotes the sign convention). But the object accelrates upwards slightly and hence the total force on the rope becomes F=(-mg)+ma Where a is the acceleration of the frame. You’ll get your required answer.

If the mass did not have a rope attached to it the force on it would be $50 g$ where $g$ is the gravitational field strength $g = 9.8 \, \rm N\, kg^{-1}$.

So using Newton's second law $F=ma$ gives

$50 \times 9.8 = 50 \times a \Rightarrow a = 9.8 \,\rm m\, s^{-2}$

so the acceleration $a$ of the mass would be $9.8 \,\rm m\, s^{-2}$.

You are told in the question that with the string the acceleration of the mass is less than $9.8 \,\rm m\, s^{-2}$ and so the net force on the mass must be less than $50\,g$ and equal to $50\,g - T$ where $t$ is the tension in the rope.

Using Newton's second law should enable you to find the tension in the rope.

We do know that the net force equals mass times the acceleration. Now if you take the block as a system, meaning you make use of the equation, mass times acceleration equals net force, on the block. Now the last element of the rope, is pulling the block upwards. This is because the block pulls the rope down, hence due to newton's third law the rope too should pull the block, but in opposite directions, that is upwards. Hence the tension acts upwards on the block. Therefore the net force is weight downwards and tension upwards, this equals mass times acceleration, hence proved. :)