Need Help (what did i do wrong)

[BunDa4Th]

An object with mass m1 = 5.00 kg, rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass m2 = 9.0 kg, as shown in Figure P4.30. Find the acceleration of each object and the tension in the cable.

mass m1 m/s^2
mass m2 m/s^2
tension N

Okay, I did a similar problem to this but the only difference was m1 and m2 location and an incline. I used this formula which I believe is correct to use here but end up with the wrong answer.

T = M_2g + M_2a and M_1g - T = M_1a i plug in T and solve for a which gives 2.8 which is incorrect then i thought i made a mistake in my equation and tried again to get 5.8 which was also incorrect.

 

[Doc Al]

Don't use "a formula"; use Newton's 2nd law to create your own formulas. Start by identifying the forces acting on each mass in the direction of motion. (Also use a consistent sign convention, so you can use the same "a" to represent the acceleration of both masses.)

 

[BunDa4Th]

would these be the two equation

m1a1 = T + m1g and m2a1 = -T + m2g

 

[Doc Al]

Nope. For one thing, the tension force on m1 and the gravitational force on m1 are perpendicular--so you can't just add them.

 

[BunDa4Th]

well, i did some reading and came upon this:

F_y = m2a_y: T - W2 = m2(-a)
F_x = m1a_x: T - f = m1(a)

edit: Okay, I read what Doc Al posted carefully and understand it. and figure that this is the right equation.

Now i have a question which is related to the same picture but this time friction is apply and i don't even understand what it means.

m1 = 10.0 kg and m2 = 4.5 kg. The coefficient of static friction between m1 and the horizontal surface is 0.60 while the coefficient of kinetic friction is 0.30.

(a) If the system is released from rest, what will its acceleration be?
(b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system?
m/s2

 

[BunDa4Th]

what is coefficient of static friction and coefficient of kinetic friction?

Fx = T - f_k = m_1a_1 Fy = n - m1g = 0

T - u_km_1g = m_1a_1

and there is this a = (m_2g - u_km_1g) / (m_1+m_2)

i tried that but it was incorrect.

 

[Doc Al]

well, i did some reading and came upon this:

F_y = m2a_y: T - W2 = m2(-a)

Good. You can replace W2 with m2g.

F_x = m1a_x: T - f = m1(a)

What's "f"?

Now i have a question which is related to the same picture but this time friction is apply and i don't even understand what it means.

m1 = 10.0 kg and m2 = 4.5 kg. The coefficient of static friction between m1 and the horizontal surface is 0.60 while the coefficient of kinetic friction is 0.30.

(a) If the system is released from rest, what will its acceleration be?
(b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system?
m/s2

Now that friction is applied, the "f" in your second equation makes more sense.

For (a), ask yourself: Does it move?

 

[Doc Al]

what is coefficient of static friction and coefficient of kinetic friction?

Since the system starts from rest in part (a), consider static friction; for part (b) it's already moving, so kinetic friction applies.

 

[BunDa4Th]

Thanks for your help. I was able to solve this and I am now ready for the quiz next week.