Resultant Force

From: Bronwyn
Date: 1 March 1999

I tried to answer the question:

A train of mass 3 tonnes is accelerating up a track inclined at
5 degrees to the horizontal. The engine force amounts to 5kN.

If resistance forces are ignored, calculate the acceleration of the train.

Using F=ma, I got an acceleration of 5000/3000 = 1.666ms^-2.

Why is this wrong?

A diagram (right) is shown for this situation.
(Note that, for simplicity, the normal reaction force is not shown)

The weight force W has been resolved into components parallel and perpendicular to the slope (shown in blue).

The resultant force that acts to accelerate the train is (F - Wsin5°) Newtons.

You were correct to use the "F=ma" formula, but the "F" in the formula is the resultant force, not the F of the diagram:

And if g = 9.8ms^-2,