# Box Pulled on Rough Horizontal Surface - Frictional Force

1. Oct 31, 2015

### physicsday

1. The problem statement, all variables and given/known data
A 100N box is initially at rest on a rough horizontal surface. The coefficient of static friction between the box and the surface is 0.6 and the coefficient of kinetic friction is 0.4. A constant 35N force is aplied to the box horizontally.

Identify from choices (a) - (e) how each change described below will affect the frictional force on the box by the surface 1 second after the horizonal force is first applied.

Compared to the case above, this change will
(a) increase the frictional force exerted on the by the surface
(b) decrease the frictional force exerted on the box by the surfact but not to zero
(c) decrease the frictional force exerted on the box by the surface to zero
(d) have no effect on the frictional force exeerted on the box by the surface
(e) have an indeterminate effect on the frictional force exerted on the box by the surface

All of these modification are changes to the initial situation shown in the diagramm.
2) The weight of the box is changed to 200N _____
3) The applied force is increased to 50N ____
4) The applied force is increased to 80N ____
5) The coefficient of static friction is increased to 0.7 ______
6) The coefficient of kinetic friction is increased to 0.5 _____
7) The coefficient of kietic friction is increased to 0.5 and the coefficient of static friction is increased to 0.7 ___
8) The weight of the box is changed to 200N and the coefficient of static friction is increased to 0.7 ___
9) The weight of the box is changed to 200N and the coefficient of kinetic friction is increased to 0.5 ____

2. Relevant equations
F(sf) = 0.6 * Fnormal = 0.6 * 100 = 60N for initial situation
F(kf) = 0.4 * Fnormal = 40 * 100 = 40N for initial situation

3. The attempt at a solution
In the initial situation, the 35N force applied is less than 60N, so the box will not move, the frictional force should be 35N, in the opposite direction of the applied force

2) increase the weight will increase F(sf) to 120N, since 35N is less than 120N, the frictional force is still 35N. My answer is d, but the answer key is a???

3) The increase in applied force should increase the frictional force to 50N, although the box still will not move. My answer is a, but the answer key is d???

4) 80N is greater than 60N, so the box will move, 1 second after, the friction should be 40N, which is bigger than 35N in initial situation. My answer is a, but the answer key is d???

5) the coefficient of static friction inrease will result Fsf = 70N. but since only 35N is applied, the friction is still 35N. My answer is d, but the answer key is a???

6) the coefficient of kinetic frition increase will result Fkf = 50N, but since 35N will not even move the box, the friction is still 35N. My answer is d, but the answer key is a???

7) the increase in both coefficient does not change anything like in question 5&6. the frictionn is still 35N. My answer is d, but the answer key is a???

8) this is combined question 2& 5, the friction is still 35N. My answer is d, but the answer key is a???

9) this is combined question 2&6, the friction is still 35N. My answer is d, but the answer key is a???

I am not sure where I am thinking wrong. Any help will be appreciated.

2. Oct 31, 2015

### Simon Bridge

In 2 you write that the increase in weight will increase the friction force... yet you answered "d" that the friction force is not affected.
Your stated reason seems to be that the speed is not affected... still being zero.
But the question is not asking about speed.
The rest of your answers have a similar mistake.

3. Oct 31, 2015

### Mister T

I agree with you. F(sf) = 120 N, it's the maximum value of static friction. The actual value is 35 N because the applied force is 35 N.

I think you've outsmarted whoever it was who made up the answer key. Unless you've somehow stated the problem incorrectly.

Last edited: Nov 1, 2015
4. Nov 1, 2015

### physicsday

Here is the original question. It is from TIPERs Sensemaking Tasks for Introdutory Physics. I am really puzzled by the answer keys. Thanks for answering my question.

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5. Nov 1, 2015

### Mister T

Is the answer key also from the TIPERs book, or did your teacher give you those answers? I will try to remember to look in my copy of the book on Monday when I'm at work. If there's a mistake in the answer key the authors need to know.

6. Nov 1, 2015

### physicsday

I think the answer is from the TIPERs book solution. I would appreciate if you could double check the answer key and let me know. Thanks again for your help!

7. Dec 8, 2015

### Mister T

The answer key I have was distributed to workshop participants and is dated November 13, 2008. It lists the answer to Numbers 2 through 9 as "d-35N". So it appears you are correct and whatever answer key was given to you is an older version or a version prepared by a confused teacher. If it was your teacher who was confused have mercy. No one is perfect.

I know that there are lots of versions out there, the most recent one I have was officially published in 2015 but doesn't appear to have this particular question in it. It does have a variant of it, though.

8. Dec 9, 2015

### Simon Bridge

OK quite a lot of time has passed so ... off the top of my head:
So the max friction force will be 60N. 35<60 so the box remains stationary, and the friction force is 35N opposing the applied force.

The max possible static friction is now 120N but the applied force is unchanged so the actual static friction is still 35N. (d)
The thing to remember about static friction is that it is equal to the (parallel/horizontal) applied force as long as the object is not moving wrt the surface.
The coefficient times contact force gives you the max force that friction can exert without the object starting to slide.

(but the weight is still 100N) 50<60 so the box is still stationary - static friction is increased to 50N (a)

80>60 so the box is accelerating though v=0 at t=0. The friction force is 40N when it was 35N before - hence the friction force has increased. (a)

(But applied force still 35N) No effect on friction force... box is still stationary with friction 35N (max static friction becomes 70N, 70>35)

(But all else unchanged) No effect - changing kinetic friction does not effect the situation for a stationary box.

... still no effect. Box is stationary so kinetic friction coeff changes are irrelevant, and increasing the max static friction still means the box is stationary so the friction force remains the same magnitude as the applied force.

Unchanged. The max static friction is doubled but the applied force is the same so the actual friction exerted on the box is unchanged.

Unchanged. Changed to kinetic coeff irrelevant - otherwise same as above.

Lets take another look at the possibilities:
(a) friction increases: if the applied force increases but less that the max static friction
(b) friction decreases but non-zero
i. if the applied force decreases, but not to zero
ii. applied force is bigger than max static friction and kinetic friction less than 35N
(c) decreses to zero if applied force zero, and/or kinetic and static friction coeffs zero
(d) no effect as discussed above
(e) indeterminate effect when something which has an ambiguous effect on friction is menmtioned ... temperature increases for eg. This may make the surface stickier or slipperier so more information is needed to make a determination.

Note: still off the top of my head so there may be exceptions and clarifications that you or others may think of.