# Thread: Mechanics/Forces Question

1. ## Mechanics/Forces Question

Can someone please help me with the following question, I need to make sure I'm using the right method.

Two trucks, A and B both weight 10,000 Kg and are attached together by a light horizontal coupling. The accelaration of the system is 0.2 ms^-2 and the resistive forces on trucks A and B are 500 N and 300N respectively. Find PN (the driving force).

So I did F= ma +800
F= 20,000*0.2 +800
F= 4800N

Now however I am told that the resistive force on truck A has increased to 2000N but PN has remained the same and I have to find the new acceleration. How would I work this out exactly?

2. So you found that the initial setup had a force of 4800N pushing it, was resisted by 800N and so exhibited an acceleration of 0.2 m/s^2 since the total mass was 20,000kg.

Now the resistive force has increased by 1500N (was 500, is now 2000). So the force of 4800N is now fighting against 2300N (=800+1500). So there's a net force of 2500N to push 20000kg, giving you an acceleration of 2500/20000 m/s^2.

3. Originally Posted by Unknown?
Can someone please help me with the following question, I need to make sure I'm using the right method.

Two trucks, A and B both weight 10,000 Kg and are attached together by a light horizontal coupling. The accelaration of the system is 0.2 ms^-2 and the resistive forces on trucks A and B are 500 N and 300N respectively. Find PN (the driving force).

So I did F= ma +800
F= 20,000*0.2 +800
F= 4800N

Now however I am told that the resistive force on truck A has increased to 2000N but PN has remained the same and I have to find the new acceleration. How would I work this out exactly?
the new total resistive force becomes 2000+300=2300
now just use the formula F=ma+2300
since driving force is same.
4800=20,000*a+2300 Find a.