Net force exerted on the two-particle system

[Demix500]

net force exerted on the system quick question- halllp lol.

Homework Statement

The vector position of a 3.10 g particle moving in the xy plane varies in time according to the following equation.

r1 = (3i+3j)t + 2jt^2

At the same time, the vector position of a 5.15 g particle varies according to the following equation.
r2= 3i-2it^2 -6jt

For each equation, t is in s and r is in cm. Solve the following when t = 2.00

(e) Find the net force exerted on the two-particle system.
i μN
j μN

Homework Equations

f=ma

The Attempt at a Solution

Took the 2nd derivative to get the acceleration vectors.

r1 : 4 j (mass is 3.1g)
r2 : -4 i (mass is 5.15g)



I tried this:

( (m2)(4 i) ) / (m1 + m2)

and got -1.93515 in the i-hat direction. It is incorrect : "Your answer is off by a multiple of ten."

 

[Doc Al]

Took the 2nd derivative to get the acceleration vectors.

r1 : 4 j (mass is 3.1g)
r2 : -4 i (mass is 5.15g)

Looks good.

I tried this:

( (m2)(4 i) ) / (m1 + m2)

Not sure what you were going for here.

What's the net force on each mass?

 

[Demix500]

Well the vector r2 is the only mass going in the i-hat direction. And since its a system, I divided by the total mass

 

[Doc Al]

Well the vector r2 is the only mass going in the i-hat direction.

Were you trying to find one component of the net force?

And since its a system, I divided by the total mass

Does that make sense? Check units.

 

[Demix500]

The answers they want is in μN. Newtons I understand, but what is μ doing in there?

 

[Doc Al]

The answers they want is in μN. Newtons I understand, but what is μ doing in there?

That just means micro = 10^-6. (Realize that you're given measurements in cm and grams.)