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.)

 

[Moetasim]

I would first check the units in the given equation to ensure they are consistent. The coefficient of static friction is a unitless quantity, while the surface normal force and weight of the object are both measured in Newtons (N). Therefore, the units do not match and the given equation is incorrect.

The correct equation for static friction is: Friction Force = coefficient of static friction * Surface Normal Force.

Using this equation, we can solve for the friction force:

Friction Force = 0.72 * (5.0 kg * 9.81 m/s^2)
Friction Force = 35.46 N

This means that a minimum force of 35.46 N is needed to start the rubber block sliding.

It is possible that the answer in the book of 106 N is for the kinetic friction force, which is the force needed to keep the object moving once it has started sliding. This force is typically larger than the static friction force.

In conclusion, the correct answer for the minimum force needed to start the rubber block sliding is 35.46 N, as calculated using the correct equation for static friction.