Breaking static friction w/ a circle

[CollegeStudent]

Homework Statement

A crate (35 kg) is located in the middle of the flat bed of a pickup truck as the truck rounds an unbanked curve in the road. The curve may be regarded as an arc of a circle of radius 80.0 m. If the coefficient of static friction between crate and truck is 0.410, how fast can the truck be moving in m/s and miles per hour without the crate sliding? (1 mile = 1609 m)

Homework Equations

The Attempt at a Solution

I'm just doing some practice problems before my midterm next week...I have NO idea how to approach this one

I have mass (35kg)
coefficient of F_S is .410

we haven't gone over something like this yet in class...but I want to practice any problems I can...

where do I even begin to start with this one?

 

[ehild]

Think: how does the crate move with respect to the ground? What force keeps the crate on the flat bed of the trunk?

ehild

 

[CollegeStudent]

Think: how does the crate move with respect to the ground? What force keeps the crate on the flat bed of the trunk?

ehild

well the crate will move depending on the movement of the truck...the truck turns too far left...the crate will move to the right

gravity is keeping the crate on the bed of the truck...and friction is (until it's overtaken) stopping the crate from moving

 

[haruspex]

If the crate does not slide, what will be its acceleration (magnitude and direction)?

 

[CollegeStudent]

If the crate does not slide, what will be its acceleration (magnitude and direction)?

if the crate does not move...there is no acceleration therefor no change in direction/magnitude...so 0

 

[haruspex]

if the crate does not move...there is no acceleration therefor no change in direction/magnitude...so 0

It doesn't move relative to the truck, but the truck is rounding a bend. The truck is moving at constant speed but its velocity (and therefore the crate's velocity) is changing. Any change in velocity is an acceleration.

 

[CollegeStudent]

okay...understood...since velocity does depend on direction...and if direction is changing so is velocity

so then if it involves a circular motion...has a radius...and a velocity

would I be using

F_c (centripetal force) = m(v ² / r)?

 

[ehild]

okay...understood...since velocity does depend on direction...and if direction is changing so is velocity

so then if it involves a circular motion...has a radius...and a velocity

would I be using

F_c (centripetal force) = m(v ² / r)?

Yes, the crate needs a force equal to the centripetal force. What force is exerted on it by the bed of the truck?

ehild

 

[CollegeStudent]

perpendicular to the surface would be the normal force...

and just for the future when I need it

Ffriction = μ_s * N

and we know μ_s is .410


but to answer your question...normal force

 

[ehild]

Yes, there is the normal force from the bed which cancels with gravity. But friction is also a force from the bed, in horizontal direction. It prevents slipping. Without friction, the crate would slip radially outward, so the force of static friction points radially inward. It supplies the centripetal force.

As you have written, the maximum static friction μ_sN, and N=mg as the crate does not accelerate vertically. And the force of static friction is equal to the centripetal force. So what is the possible maximum speed when the crate does not slip?

ehild

 

[CollegeStudent]

Yes, there is the normal force from the bed which cancels with gravity. But friction is also a force from the bed, in horizontal direction. It prevents slipping. Without friction, the crate would slip radially outward, so the force of static friction points radially inward. It supplies the centripetal force.

As you have written, the maximum static friction μ_sN, and N=mg as the crate does not accelerate vertically. And the force of static friction is equal to the centripetal force. So what is the possible maximum speed when the crate does not slip?

ehild

hmmm so what you're saying is that the force of static friction would be equal to the centripetal force

so

F_s = F_c?

μ_s N = m(V²/r)

μ_s * mg = m(V² / r)

m's cancel so

μ_s * g = V² / r

so

.410 * 9.8 = V² / 80

so V² would equal (.410 * 9.8) * 80?

and V would equal √((.410 * 9.8)*80) ?

is that correct?

 

[haruspex]

and V would equal √((.410 * 9.8)*80) ?

is that correct?

Yes, the units being...?

 

[CollegeStudent]

well let's see..

the square root...of meters/second² times meters

so meters ² / seconds ²

square root would make it m/s

so the answer I'm getting would be

17.93m/s

we want this in miles per hour so

17.93 m/s (1 mile / 1609 meters)(3600 seconds / 1 hour)

this comes out to

40.11 m/h

look right?

 

[haruspex]

Yes, that looks right. I approve of the way you embedded the units conversion in the equation. I wish more would do it like that.

 

[CollegeStudent]

Yes, that looks right. I approve of the way you embedded the units conversion in the equation. I wish more would do it like that.

that's how I've always done it...it keeps me better organized! ...but thank you for your help!