Co-efficient of Static Friction Question

[Sabellic]

Homework Statement

If the coefficient of static friction for a 5.0 kg rubber block on a floor is 0.72, how much force is needed to start it sliding?

Mass=5.0kg
Co-efficient of Static Friction=0.72
Friction=?

Homework Equations

coefficient of static friction = Friction Force/Surface Normal Force (equal to the weight of the object)

The Attempt at a Solution

0.72=Friction/(5)*(9.81)
0.72=Friction/49.05
0.72*49.05=Friction
35N=Friction

Therefore, an applied force of at least 35 Newtons is needed to overcome the Friction Force and start the object moving. But the answer in the book gave me 106 Newtons. I don't know how that was arrived at...but strangely enough 35 * 3 = 106.

Thanks in advance for clearing up the matter for me.

 

[Doc Al]

Sounds like you are correct and the book is wrong. (I assume the force is meant to be applied horizontally.)