# Newtons second law

1. Jul 9, 2007

### sweet-buds

1. The problem statement, all variables and given/known data
F=ma can be written as F=ma(along x, y or z-direction, )

when applying this equation along say horizontal x direction, the direction of acceleration needs to be taken as it is a vector.

i mean say 2 forces of 5 N and 3 N are applied on a 2 kg body that moves towards left as 5 N is applied towards left and 3N towards right. Now

According to me ,
5-3=2(-a)
= 2=-2a
=-1 m/s^2=a

this -1 indicates that accn is towards left.

I am confused with the idea should eqn be 5-3 = -2a or 5-3 =2a. i would go with the former one.

2. Relevant equations

3. The attempt at a solution

2. Jul 9, 2007

### bsmith2000

Assuming right to be positive and left to be negative, and the set-up as such:

3 N ----> [2kg] <---- 5 N

Then we have F_net = m*a = 3 N + (-5 N) = - 2 N. You have a net force of 2 N in the negative direction (meaning it will be pushed to the left). Since -2 N = m*a and m = 2 kg, then we have -1 m/s^2. Since this is negative, once again, the body will accelerate to the left.

Does this make sense?

Last edited: Jul 9, 2007
3. Jul 9, 2007

### sweet-buds

how is that possible? I would say since the 5N force is greater so the F_net=3-5
=-2 N, a force that act towards left. Now, -2=m*(-a)=>-2=2*(-a)=>1=a, as I pressume the body will move towards left due to net -2 N force and since acceleration is towards left, I have used a - ve sign before a.

My confusion arises in the second eqution , F=ma where F is the net force acting on the body and lets say the force is in negative horizontal or vertical
direction, arent we suppose to assume the direction of acceleration also negative as acceleration is vector.

4. Jul 9, 2007

### bsmith2000

Ack! I am terribly sorry - I made a sign error when I added 3 and -5. You are correct! I edited the post with the necessary changes.
You are also correct when you say "he force is in negative horizontal or vertical
direction, arent we suppose to assume the direction of acceleration also negative as acceleration is vector."

5. Jul 10, 2007

### sweet-buds

help me urgent: newton's 2nd law

You are also correct when you say "he force is in negative horizontal or vertical
direction, arent we suppose to assume the direction of acceleration also negative as acceleration is vector."[/QUOTE]

i got this problem in Haliday. There are 2 blocks connected by a rope. onr of the blocks is a sliding one and the other is a hanging one and the rope passes over a pulley and the pulley is connected at the edge of the table/surface. The sliding block moves towards the right and the hanging block down. Haliday has applied the newtons second law to the two blocks.

The sliding block has mass M and hanging block a mass of m. haliday has said the the equation for hanging block is T-mg =-ma. i understand the fact that the hanging block is moving down in -ve y directio, that's why a istaken -ve here but doesn't that mean that mg>T for hanging block to move down. I mean shouldn't the equation be mg-T=-ma?