Two blocks of masses m1= 1 kg and m2= 2 kg are connected by a massless cord passing over a massless,?

Question

frictionless pulley. the pulley is pulled up with a force F = 50 N. what are the acceleration of teh two masses with respect to the ground?

Answer 1

I will assume each mass is hanging down so all forces and motion are vertical

Again a FBD for each

m1:
T-m1*g=m1*a1
Positive is upward

m2:
T-m2*g=m2*a2

pulley
50N-2*T=mp*ap since the pulley is massless
mp=0
T=25 N
I will use 10 for g
25-2*10=2*a2
5/2=a2

25-10=1*a1
15=a1

Note that these accelerations are w/r/t the ground. Clearly the m2 is slower in the upward direction since it is pulling the m1 upward towards the pulley if you move the frame of reference to the pulley.

That was a fun one

j

Answer 2

We have to set up the equation first.

F = ma, Force equals mass times acceleration

We're trying to find acceleration, so:

a = F/m, Acceleration equals Force divided by mass

The total mass of our system is M2 minus M1, because one mass is acting against the other over a frictionless, massless pulley, so.

a = F/(m2 - m1)

F is given to be 50N, so the equation will look like:

a = 50N/(2kg - 1kg) = 50N/1000g = 0.05m/s^2

With respect to ground, m1 will be moving up with a positive acceleration 0.05m/s^2, and m2 will be moving down with a negative acceleration of -0.05m/s^2.

We could be cynical and say zero, because 50N is not enough force to overcome gravity pulling the whole system to ground. Neither mass will be able to get off the ground if they are on the ground at t = 0.

At any rate, these masses are connected to one another......how can they have different magnitudes of acceleration? Just wondering how the people above came up with different accelerations.

Answer 3

considering the system as a whole... 50 N force acts on a 3kg mass... hence the pulley moves with an acceleration of 50/3 m/s^2. now looking at each mass with respect to the pulley itself... for m2:- 2g-T=2a; and m1:- T-g=a where g is 9.8 m/s^2 and T is tension in string

hence a= g/3 i.e. 3.27 m/s^2

Therefore... acceleration of m2 is 13.40 m/s^2 upwards
and m1 is 19.92 m/s^2 upwards...

Answer 4

because the coef. of static friction is given, it truly is worth checking to confirm if the gadget will bypass in any respect .. Friction between m1 and table .. Ff = ?(s).R .. .. (R = perp. stress, m1g) Ff = 0.40 4 x (41kg x 9.80) .. .. .. Ff = 176.80 N at the same time as the gadget is at relax the stress contained in the string, T = m2.g (weight of m2) T = 15kg x 9.80 .. .. T = 147.0 N the stress isn't sufficient to bypass m1 from relax .. ?T continues to be at 147.0 N

Answer 5

dis s a case of simple atwood mach. in acc.
widout acc, Acc= m2 *g-m1/m1*g+m2 i.e 10/3=3.333metre/sec
now consider d system as a whole, tot. wt.=3kg,tot force=50N,acc of system in upward dir.=50/3=16.666
acc of m1=3.3333333+16.6666666666=20met/sec in upward dirn
acc of m2=16.6666666666-3.3333333333=13.33333333met/sec in upward dirn