# Homework Help: Acceleration of a crate

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1. Nov 9, 2015

### Jonski

1. The problem statement, all variables and given/known data
At the instant shown, the driver of the truck has just pressed the accelerator pedal down and the truck has suddenly acquired a tangential acceleration of 2.2m/s^2.
Coefficient of static friction between crate and tray = 0.4
Coefficient of kinetic friction between crate and tray = 0.3
Find the acceleration of the crate?

2. Relevant equations
a(normal) = v^2/r
F=ma
F = mv^2/r
3. The attempt at a solution
a(tangential) = 2.2m/s^2
a(normal) = 6.1^2/112 =0.33
a = 2.225m/s^2
F = 270 *2.225 = 600.8N
mv^2/r = -N + m.g.cos(15)
89.7=-N+2555.84
N=2466.1N
μN = 986.4N
986.4 > 600.8
Box doesn't move

However this is not the correct answer as I know it moves.
Thanks

Last edited: Nov 9, 2015
2. Nov 9, 2015

### haruspex

In the statement of the question you only mentioned finding the acceleration of the truck, which you did. So I assume the next part is asking whethe the crate slides.
Keep the tangential and radial components of the truck's acceleration separate. Draw a free body diagram of the crate. What forces act on it parallel to the slope and what forces perpendicular to the slope? If it does not slide, what is its acceleration in each of those directions? What equations does that give you?

3. Nov 9, 2015

### Jonski

Well in the perpendicular direction you have the normal force, weight force*cos(15) and centripetal force
In the parallel direction you have friction force and weight force*sin(15) @haruspex

4. Nov 9, 2015

### haruspex

Almost right.
In nonlinear motion, you can choose to think in terms of centrifugal or centripetal.
If you choose centrifugal then it is a ("fictitious") applied force, i.e. it comes into the $\Sigma F=ma$ equation like any other applied force.
If you choose centripetal then it is a resultant force, not an applied force. Specifically, it is that component of the resultant force that produces the centripetal acceleration: $\Sigma F_{radial}=F_{centripetal}=ma_{centripetal}$.
So, write out those equations.

5. Nov 9, 2015

### Jonski

So Fcentripetal = mv^2/r = 270*6.1^2/112=89.7N
ΣF = mgcosΘ - N = 2555.84 - N = 89.7N
N = 2466.14N

Is that right? Not too sure what you mean

6. Nov 10, 2015

### haruspex

Yes, good so far.
I see you've updated post #1 to reflect this. But you are missing a force in the tangential direction.

7. Nov 10, 2015

### Jonski

Do you also have something to do with the acceleration of the truck? Otherwise I don't know

8. Nov 10, 2015

### haruspex

Hint 1: Forget the curvature of the hill and truck acceleration for a moment. Suppose the truck were going up a steady incline at constant speed, would some friction still be needed to keep the crate in place?
Hint 2: Whenever you write force equations that involve, say, $F\sin(\theta)$ for some force F and angle theta, consider that in some other equation you might expect to see a term like $F\cos(\theta)$.

9. Nov 10, 2015

### Jonski

Some friction would still be needed otherwise the crate would fall of.
Also i think the force and angle thing you're talking about is the weight force, which I included in both equations

10. Nov 10, 2015

### haruspex

You mentioned it in post #3, but I still don't see an equation involving mg sin(theta).

11. Nov 10, 2015

### Jonski

So if F=ma=270*2.225=600.8N
Then the N = mgcosΘ - mv^2/r = 270*9.8*cos(15) - 270*6.1^2/112=2466.1N
So if it were to stay with static friction:
μN*sinΘ=0.4*2466.1*sin(15)=255.3N
Since this is less than 600.8N we know it moves.
Now we not its slips and governed by kinetic friction
F = μNsin(15)=0.3*2466.1*sin(15) = 191.5N
Since F = ma, a = F/m = 0.71m/s^2
Is that right?

12. Nov 10, 2015

### haruspex

What is the relevance of μN*sinΘ?
You are looking (I hope) at forces tangential to the slope. The maximum friction, μN, is tangential to the slope. μN*sinΘ would be the vertical component of that.
What other force has a component tangential to the slope? Reread my posts #8 and #10.