Regarding tension force

Question

In FBDs of a pulley system, we usually consider the string to be massless, meaning the tension force acting anywhere on the string is the same.

Mass------->(tension)-------(tension)<---------pulley

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Bigger mass

Assuming that the bigger mass has a net acceleration downwards, it must be pulling the smaller mass towards the right.

Also, the tension forces in the upper string are equal and opposite, implying that they must cancel eachother.

So my question is- what causes the horizontal motion of the smaller mass?

Also, there is a high possibility that I'm completely wrong, if so please guide me otherwise.

Thank you.

Answer 1

The tension in the string pulls up on the mass hanging down so that the net force on that mass is F_net = Mg - T. The tension force T is pulling on the mass on a frictionless table giving it an acceleration a = T/ m or T = ma. The T can be substituted into F_net = Mg - T giving

F_net = Mg - ma. F_net = Ma, so Ma = Mg -ma. the acceleration of both masses is then

a = Mg / (M + m).

My college physics text states that the pulley reverses the tension in the string.

The diagram should be

Little mass ----------> pulley

↑

↑

↑

big mass