# Physics Quiz 4

*Figure1 is a top view of a block sitting on a table.

The block weighs 10 Newtons.

The coefficient of static friction between the block and the table is 0.8 (the same in all directions).

A force F1 = 6 Newtons is applied to the block (yellow arrow).

A force F2 (blue arrow) is then slowly applied in the direction shown. At what magnitude of F2 does the block begin to slide?

*  2*SQRT(7) Newtons

Thank you.

F2 >= 5N and some change.

Work
[spoiler]
sqrt(6^2 + F2^2) >= 10 * 0.8
or
6^2 + F2^2 >= 64
or
F2^2 >= 28
or
[/spoiler]

*Make the following changes to the original problem:

• The static coefficient of friction in the “Y” direction (the direction of the yellow arrow) is mu_y = 0.8

• The static coefficient of friction in the “X” direction (the direction of the blue arrow) is mu_x = 0.2

• The static coefficient of friction varies linearly with theta from mu=0.2 in the “X” direction to mu=0.8 in the “Y” direction as theta goes from zero (X direction) to pi/2 (Y direction).

Now find the value of Fx at which the block begins to slide.

## f2^2+36=(10(mu))^2 Sqrt[f2^2+36 [spoiler]=10(mu) Sqrt[f2^2+36]/10=mu

[spoiler=]Sqtr[f2^2+36]/10=(1.2arctan(6/f2))/pi+.2
Sqrt[f2^2+36]pi=12arctan(6/f2)+2pi
Sqrt[f2^2+36]-12arctan(6/f2)=2pi
::ouch:: that’s ugly
Wolfram Alpha couldn’t make it look any better, but it gives us an approximation for f2 of about [spoiler=]10.8N[/spoiler]
[/spoiler][/spoiler]

You were doing great until you dropped the factor of “pi” on the left hand side in the very last equation. The numerical answer you got should have been a red flag that something was wrong.

Fix that, get a new numerical value for F[sub]2[/sub], then ask yourself “what happens when F[sub]2[/sub] = 2 Newtons?”

**

Can you list the plane on which the block rests, as well as the direction of the gravitational force?

F = sqrt(36 + F2^2)
F = 10*(.2 + .6 (theta/(pi/2)))
theta = arctan(6/F2)
F = 10*(.2 + .6 (2arctan(6/F2)/pi))
sqrt(36 + F2^2) = 2 + 6 (2arctan(6/F2)/pi)
F2 ~= 2.491 N

I’m confused; does this mean that it takes MORE sideways force to move the block than it would if there was no vertical force? For example, wouldn’t the block move if F1 = 0N and F2 = 2N?

Its not needed to be specified. It has already been given sort of. If the block is sitting on a table, we assume the plane is parallel to the ground and perpendicular to the gravitational force. Remember, it is a top view of the block.

Jason

Excellent.

I’m confused; does this mean that it takes MORE sideways force to move the block than it would if there was no vertical force? For example, wouldn’t the block move if F1 = 0N and F2 = 2N?

Excellent. You have asked the \$64K question.

If you assume that the coefficient in any given direction is not increased by force components not in that direction, then the answer is that the block will start to slip when F[sub]2[/sub]=2 Newtons. See attached “solution.pdf”.

However, these 2 assumptions:

1: linear change in mu for angles between 0 and pi/2, and

2: mu is not increased by force components not in direction being considered

are I think open questions.

In FRC, there are wheels which are said to have different mu in the forward and sideways directions. I have looked but never seen any test data to show what happens when force is applied at a variety of angles between 0 and pi/2.

*

solution.pdf (42.7 KB)

solution.pdf (42.7 KB)

Top view with respect to what, though? The table, or the plane the forces are acting on? The ambiguity of the question leaves it perfectly open to interpretation that the table is flush with any of the six faces of the block. Assumptions are useful in solving situations like this, but they can also lead to highly erroneous answers if they are wrong.

Its better that I make that mistake here than on the AP exam:o

The plane on which the forces are acting and the table are the same plane, the normal force is perpendicular to that plane, the direction of gravity, irl, is down with respect to an observer in the same gravitational field as the object in question. since we are looking from the top, down can be assumed to be away from us.

Agreed. Try solving such problems without making assumptions though. I’ve never had a professor explicitly spell out every detail of a problem. They always leave it up to the student to make a few assumptions. In a real world example, you never know all the details. Just learn what you can and do your best from there.

Jason

Furthermore, in the ‘real world’, half the time we don’t know what assumptions we’re even making, yet somehow the work has to get done.

See attachments.

Plotting the “linear” model of μ vs θ in polar coordinates suggests it may not be a good model.

Perhaps an elliptical interpolation between μ[sub]x[/sub] and μ[sub]y[/sub] would be a better model. See attachments.

I haven’t done the math yet, but I think the elliptical model will solve the “2 Newton paradox” discussed earlier in this thread.

*

Solution is 1.3 Newtons using the elliptical interpolation. See attachments.

*

Very interesting thread. Here is some neat related tire info:
http://www.insideracingtechnology.com/tirebkexerpt2.htm

This would make an excellent science fair project.