Resultant Force

  • From: Bronwyn
  • Date: 1 March 1999
  • Subject: F=ma

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?

Maths Help suggests:

Diagram showing Weight W and Engine Force F

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:

acceleration = resultant force / mass

And if   g = 9.8ms-2,

a = 0.813ms^-2 (3sf)
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