# Applying Force On An Angle

1. Feb 4, 2007

### danago

1. The problem statement, all variables and given/known data
A lawnmower weighing 20kg has a handle attached at 35 degrees to the horizontal. If a man wishes to push the lawnmower, so that after 2.5s, its velcoity is 2 m/s, what force should he apply along the handle.

2. Relevant equations
$$\sum {\overrightarrow F } = m\overrightarrow a$$

3. The attempt at a solution
ok, since after 2.5 seconds, it requires a velocity of 2 m/s, i found the acceleration of the mower to be 0.8 m/s/s. Since it has a mass of 20kg, the net force applied must then be:

$$\sum {\overrightarrow F } = 20 \times 0.8 = 16N$$

So the horizontal component of the force he applies must be 16N in the direction of acceleration.

The horizontal component of the force (F) he applies down the handle will be $$F cos 35$$. So solving for F:

$$Fcos35=16$$

Gives F=19.5N.

According to the answer book, the answer is 120N. Where have i gone wrong?

Thanks,
Dan.

Last edited: Feb 4, 2007
2. Feb 4, 2007

### Hootenanny

Staff Emeritus
If the question is quoted correctly, then I would agree with your answer. Does the text not mention any other factors, such as friction?

3. Feb 4, 2007

### danago

Nope, no friction at all. What ive said is pretty much quoted word for word from the book.

4. Feb 4, 2007

### Ja4Coltrane

yeah the book is def. wrong

5. Feb 5, 2007

### danago

ok thats good to hear :) Thanks for the help