Acceleration needed to keep an object from sliding down a windshield

[pochemuchka]

Homework Statement

You are driving at a reasonable constant velocity in a van with a windshield tilted 120∘ relative to the horizontal. As you pass under a utility worker fixing a power line, his wallet falls onto the windshield. Assume the coefficient of static friction between the glass and the wallet to be 0.50.

Relevant Equations

F=ma, N=-mg, Fg=mg, Fs=(mu)N

I think I am on the right path but I'm having trouble cancelling out the mass of the wallet. I drew a free body diagram and know that there are three forces acting on the wallet: gravity, friction, and normal force (including the van's acceleration).

The y-component of the friction and normal forces should cancel the force of gravity. I have the equation as mg = sin60(mu)(N) + sin30(N) [N is normal force].

I don't know how to cancel out the mass so I can use the normal force to find acceleration in the x-direction. Any help?

 

[kuruman]

Can you find an expression for the normal force in terms of the weight? Then the mass will cancel out. Also your equation seems to be incorrect. What is the right hand side supposed to express? You need to get two equations from your free body diagram, one for the vertical direction and one for the horizontal direction.

 

[wrobel]

I think first you should stop and return the wallet to the owner

 

[pochemuchka]

Can you find an expression for the normal force in terms of the weight? Then the mass will cancel out. Also your equation seems to be incorrect. What is the right hand side supposed to express? You need to get two equations from your free body diagram, one for the vertical direction and one for the horizontal direction.

Normal force usually is -mg, but in this case it will be -mg+the force applied on it by the accelerating window. The right hand side is the y-components of the frictional force and normal force.

For the x-direction, my equation would be Fnet=ma, with Fnet being the forces in the x-direction. This would be the cos30(Fnorm) - cos60(mu)(Fnorm) = ma.

I guess my difficulty is in eliminating the Fnorm or relating it to mass in some way.

 

[kuruman]

I think first you should stop and return the wallet to the owner

No, OP first needs to solve the problem and calculate where the wallet will land after the van stops.

 

[pochemuchka]

No, OP first needs to solve the problem and calculate where the wallet will land after the van stops.

Any help on the previous reply?

 

[kuruman]

Any help on the previous reply?

Suppose you wrote your two equations with one above the other like so
cos30(Fnorm) - cos60(mu)(Fnorm) = ma
sin60(mu)(Fnorm) + sin30(Fnorm) = mg
Suppose you divide the top equation by the bottom equation. What happens?

 

[pochemuchka]

Suppose you wrote your two equations with one above the other like so
cos30(Fnorm) - cos60(mu)(Fnorm) = ma
sin60(mu)(Fnorm) + sin30(Fnorm) = mg
Suppose you divide the top equation by the bottom equation. What happens?

UGH I can't believe I didn't see that. The masses and Fnorm cancels out, giving me tan30-tan60=g/a.

Thanks so much!

 

[neilparker62]

Alternatively you could resolve forces parallel and perpendicular to the 'inclined plane' formed by the windscreen:

$$ mgsin(60) - macos(60) = μ(mgcos(60) + masin(60)) $$

I am just trying to get my head around -macos(60): had initially wanted to write +macos(60) but then the answer didn't quite agree per Kuruman's post #7 above! I think it is because the window slipping underneath the wallet gives it an apparent acceleration up the slope.

Btw am not sure if your simplification tan30 - tan60 = g/a is correct (?)

 

[kuruman]

I am just trying to get my head around -macos(60): had initially wanted to write +macos(60) but then the answer didn't quite agree per Kuruman's post #7 above! I think it is because the window slipping underneath the wallet gives it an apparent acceleration up the slope.

I am not sure I understand what you mean by "apparent acceleration". The question of the problem is not included in the statement posted by @pochemuchka. I assumed that it was something like "Find the van's acceleration so that the wallet barely stays on the windshield without sliding down". This requires a force of static friction up the windshield. If the question were "Find the van's acceleration so that the wallet barely stays on the windshield without sliding up", the static friction would be directed down the windshield and the answer would be obtained from the previous answer simply by changing the sign of $\mu.$ I assumed that the problem wants one to find the smaller of the two possible accelerations.

Btw am not sure if your simplification tan30 - tan60 = g/a is correct (?)

It is not correct. If that were true, then$$\frac{12+6}{2-1}=\frac{12}{2}-\frac{6}{1} =6-6=0.$$The correct result (obtained without the use of a calculator) is 18.

 

[neilparker62]

Suppose you wrote your two equations with one above the other like so
cos30(Fnorm) - cos60(mu)(Fnorm) = ma
sin60(mu)(Fnorm) + sin30(Fnorm) = mg
Suppose you divide the top equation by the bottom equation. What happens?

$$ \frac{a}{g}=\frac{cos(30)-μcos(60)}{μsin(60)+sin(30)}⇒a=g(5\sqrt{3}-8)$$Doesn't give a=18 m/s²?

 

[kuruman]

$$ \frac{a}{g}=\frac{cos(30)-μcos(60)}{μsin(60)+sin(30)}⇒a=g(5\sqrt{3}-8)$$Doesn't give a=18 m/s²?

It does not. Try the calculation again at the very last step.

 

[neilparker62]

I am not sure I understand what you mean by "apparent acceleration".

I'm not sure that I do either - best I can come up with for an acceleration up an accelerating windscreen!

Working back from the equation I wrote to an FBD, I obtain the following - don't know if it makes sense ? Car's acceleration produces a force pushing the wallet up the windscreen - I'm seeing this as the parallel component of reactive force acting on the wallet.

 

[neilparker62]

It is not correct. If that were true, then$$\frac{12+6}{2-1}=\frac{12}{2}-\frac{6}{1} =6-6=0.$$The correct result (obtained without the use of a calculator) is 18.

Sorry - I misunderstood you. 18 is the correct answer to the simple example you gave. I was thinking you meant the correct answer to the OP's problem. Duh!

 

[kuruman]

I'm not sure that I do either - best I can come up with for an acceleration up an accelerating windscreen!

You free-body diagram is drawn in a non-inertial frame (the POV of the driver). In that case, you bring $m\vec a$ inside the FBD with a change in sign and treat it as a "fictitious" force. Of course, in the non-inertial frame the sum of the forces. real and fictitious, must be zero. Since most FBDs are drawn in an inertial frame, when you draw one in a non-inertial frame it is helpful to clarify your choice.

 

[kuruman]

Sorry - I misunderstood you. 18 is the correct answer to the simple example you gave. I was thinking you meant the correct answer to the OP's problem. Duh!

My mistake for not clarifying whom to whom my post was addressed.