Engineering Statics. Friction involving two bodies.

[thepatient]

This here is actually an exam question from exam taken last week. I got near full credit for it, but I didn't do this problem properly. Even though I got a very high score, I feel I wasn't deserving of it.

Haha, anyway, I still was a bit unsure of why the assumptions were made in the problem. Maybe someone here can help me understand why such assumptions should be done.

Homework Statement

A board is pushed over a sawhorse slowly by a carpenter. Board has uniform weight of 3 lb per linear foot. Saw horse weighs 15 lb. Determine if the saw horse stays in position, slips or tips when the board is pushed forward and d = 14 ft.

Homework Equations

Net force = 0
Net moment (torque) = 0
Maximum static friction = mu(s) * Normal

The Attempt at a Solution

First, considered the board as a free body. On the free body diagram I included horizontal force of push, normal force upward at end of d, friction force to left at end of d, and weight equal to length times weight per linear foot.

Since the problem said horizontal push, I assumed there was only one component of force and no upward component. The correct answer had an upward force where the carpenter pushes, so that was one mistake.

Then considering the saw horse by itself, I had two forces of friction on the legs to the left, two normal forces upwards. Weight downward at center, force due to board downward at center of saw horse, and friction to the right due to board at the top.

According to the solution, you have to consider the case when the saw horse slips and when it tips, which makes sense. I made a bad assumption and said the saw horse tips if the normal force on the left leg is zero or less than zero, and slips if the normal is more than zero. Or maybe it is a correct assumption, but not enough to solve the problem.

The solution said that if the object slips, then force Px (pushing force off the man) equals the force of friction on base of saw horse, which is equal to .3 times the net downward force. I didn't understand why it didn't consider friction on both legs though. (I'll post solution as well).

Basically, I just wasn't sure why just one leg had a force of friction and normal force in both cases. Also on free body diagram of the case where the saw horse slips, there is no horizontal force of friction at the top due to the board. Maybe its just a typo. Thanks. :]

 

[NascentOxygen]

Then considering the saw horse by itself, I had two forces of friction on the legs to the left, two normal forces upwards. Weight downward at center, force due to board downward at center of saw horse, and friction to the right due to board at the top.

Friction to the left, I think, because he seems to be pushing the plank towards the right.

The solution said that if the object slips, then force Px (pushing force off the man) equals the force of friction on base of saw horse, which is equal to .3 times the net downward force. I didn't understand why it didn't consider friction on both legs though. (I'll post solution as well).

When all the "weight" is carried by the legs on one side of the sawhorse, it will either tip or--if it's not already slipping--it will start to slip.

Basically, I just wasn't sure why just one leg had a force of friction and normal force in both cases.

I suppose both carry some vertical load. But whether you consider one leg carrying some of the load (and its coefficient of friction) and the other leg carrying the remainder of the load (and its coefficient of friction), the result will be identical with considering one leg to carry all of the load (at that same coefficient of friction).

 

[JHamm]

I would imagine friction was considered on only one leg since friction is independent of surface area, only a guess though.