Solving Angle Pulling Force w/ Friction & Mass

[dalcde]

Homework Statement

A box of mass 30kg is being pulled along rough horizontal ground at a constant speed using a rope. The rope makes an angle of 20 degrees with the ground. The coefficient of friction between the box and the ground is 0.4. The box is modeled as a particle and the rope as a light, inextensible string. The tension in the rope is P Newtons.

Homework Equations

The Attempt at a Solution

The normal force is
$$30g-P\sin 20^\circ.$$
The force of friction is
$$(30g-P\sin 20^\circ)0.4,$$
which should be equal to the pulling force,
$$P\cos 20^\circ.$$
Hence
$$(30g-P\sin 20^\circ)0.4=P\cos20^\circ.$$
Taking g as 9.8 and solving the equation yields
$$P=109,$$
corrected to 3 significant figures.

However, the answer says that it should be 125 instead of 109. What did I do wrong? Also, the book seems to prefer to have 3 significant figures but 9.8, the numerical value of g, only has 2, which seems weird to me. Any justifications for it?

(Sorry, I don't know how to type the degree sign in LaTeX. Can anyone teach me how to?

EDIT: added the degree signs

 

[rock.freak667]

I would say that you are correct. Even putting g = 10 N/kg would not get it to 125 N.

 

[SammyS]

...
(Sorry, I don't know how to type the degree sign in LaTeX. Can anyone teach me how to?

20^\circ

$$(30g-P\sin 20^\circ)0.4=P\cos20^\circ$$

 

[Fewmet]

I would say that you are correct. Even putting g = 10 N/kg would not get it to 125 N.

I am also getting 190 N.

You can also do degrees by making a lower case o into a superscript with the x² button at the top of the message composing window: cos20^o