Question about setting up freebody diagrams

[canadianofpeace]

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When you set the system equal to ma, is ma ever negative?

For example in a pulley system, imagine mass 1 (light) which is being pulled up and mass 2 which is going down (heavy)

For the first system the forces are
Tension (Up), Gravity (Down). So I set up the quation T-mg=ma

For the second system the forces are also
Tension (Up), Gravity (Down). However, acceleration here is also down. My notes indicate that when you assign up and down as positive and negatives, accelerations also follow. So I set up my equation like this

T-mg=-ma

However, the solution manual says it is supposed to be T-mg=ma (same as before?)

A bit confused and would appreciate help

 

[gettheleadout]

For the second system the forces are also
Tension (Up), Gravity (Down). However, acceleration here is also down. My notes indicate that when you assign up and down as positive and negatives, accelerations also follow. So I set up my equation like this

T-mg=-ma

However, the solution manual says it is supposed to be T-mg=ma (same as before?)

For the second system T < mg and so where ∑F = T - mg you will find that ∑F < 0. You are right that the acceleration will be negative, and that's what you'll find by referring to Newton's Second Law: ∑F = ma. We know ∑F is a negative number, so let's say ∑F = T - mg = –x. Substitute the value in to solve for a:

–x = ma
a = – (x/m)

You have a negative acceleration. You don't need to modify the equation because as long as you put signs in properly you get signs out properly.