physics lab question.... in need of help... atwood machine?

Question

the lab says to calculate teh estimated frictional force f using a=(m2-m1)g-f / (m1+m2) then you have to change the problem into f=(m2-m1)g-(m1+m2)a

i dont know how to find the acceleration so that i can solve the problem... can someone help me?

Answer 1

Consider a FBD of the m1 as the larger mass

T2-m2*g=m2*a
and
looking at m1
m1*g-T1=m1*a

and the pulley
T1=T2+f
combine and simplify
f=T2-T1
T2=m2*g+m2*a
T1=m1*g-m1*a

f=(m2-m1)*g+(m2+m1)*a
That's different from what you have
in mine
a=(f-(m2-m1)*g)/(m2+m1)

In order to calculate a you need to do some measurements.

Find a position versus time plot of either mass starting from rest and you will have:
s(t)=s0+.5*a*t^2
plug in your data and calculate a

j

Answer 2

My answer for a million would have been "no action", whether it relies upon on what you're interpreting in Physics. 2 is definitely soliciting for a particular determine, i think of. purely undergo in suggestions that acceleration because of gravity is often 9.8m/s^2. devoid of stress resisting this pull, acceleration would be equivalent to that variety. 3 has no longer something to do with the rope. each and every mass is suspended over a pulley, and so one pulls straight away on the different for the period of the string's rigidity. The stress each and every mass exerts is comparable to the stress of gravity on the plenty, needless to say. So F = mg. g is continuous, so it is equivalent for the two plenty, and we are informed that each and every mass is equivalent, so F is likewise equivalent for the two plenty. So because of the fact that F = ma, and m is equivalent for the two plenty, and F is equivalent for the two plenty, a ought to for this reason be equivalent besides. temporarily: it is purely because of the fact the two plenty are exerting equivalent and opposite forces on one yet another, so the forces (and for this reason, the accelerations) cancel out.