Thread: Mechanics Equlibrium Involving a Slope - Faulty Answer?

I've just completed a question in the 'practice exam paper' section of my textbook, to find that I have the answer wrong. However, I can't see how the answer given can be right.

Here's the question:
A suitcase of mass 16KG is placed on a ramp inclined 15 degrees to the horizontal. The coefficient of friction between the suitcase and the ramp is 0.4. Determine whether the suitcase rests in equilibrium on the ramp, and state the magnitude of the frictional force on the suitcase.

My working is as follows:
Friction = 0.4 * R (normal contact force)
Friction = 0.4 * 9.8 * 16 * cos(15) = 60.58N (2dp)
Weight parallel to slope = 9.8 * 16 * sin (15) = 40.58N(2dp)

The answers in the back of the book state that the case is in equilibrium with a frictional force of 40.6N. I fail to see how this can be so, with a frictional coefficient of 0.4. Do you think the answers are incorrect or is there something i'm missing?

Help would be appreciated!

Originally Posted by StaryNight

I've just completed a question in the 'practice exam paper' section of my textbook, to find that I have the answer wrong. However, I can't see how the answer given can be right.

Here's the question:
A suitcase of mass 16KG is placed on a ramp inclined 15 degrees to the horizontal. The coefficient of friction between the suitcase and the ramp is 0.4. Determine whether the suitcase rests in equilibrium on the ramp, and state the magnitude of the frictional force on the suitcase.

My working is as follows:
Friction = 0.4 * R (normal contact force)
Friction = 0.4 * 9.8 * 16 * cos(15) = 60.58N (2dp)
Weight parallel to slope = 9.8 * 16 * sin (15) = 40.58N(2dp)

The answers in the back of the book state that the case is in equilibrium with a frictional force of 40.6N. I fail to see how this can be so, with a frictional coefficient of 0.4. Do you think the answers are incorrect or is there something i'm missing?

Help would be appreciated!

yes, there is ...

friction is smart ... it only does what it has to do to maintain equilibrium.

remember that the formula for static friction is an inequality ...

, reaching equality at

the frictional force 0f 40.6 N makes the net force 0 ... static equilibrium.