Calculating the Minimum Acceleration for a Wooden Block to Stay on a Cart

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SUMMARY

The discussion centers on calculating the minimum acceleration required for a 200 g wooden block to remain stationary on a 10 kg cart with a coefficient of static friction of 0.30. Participants clarify that the block's weight (1.96 N) and the friction force (0.59 N) must be considered to determine the cart's acceleration. The correct approach involves setting the friction force equal to the mass of the block multiplied by the cart's acceleration, leading to a calculated minimum acceleration of 45.7 m/s². The cart's mass is not directly needed in the final calculations, as the focus is on the forces acting on the block.

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frigid
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Maybe it's just me, but i would classify this one as hardcore...

Due to a cart's sudden acceleration, a 200 g wooden block remains on the front of the cart and does not slide or fall off.

The mass of the cart is 10 kg. The coefficient of static friction between the block and the front of the cart is 0.30

Calculate the cart's minimum acceleration if the block is to remain in place


i'm not too sure what to do with this one

I understand that the rapport between the maximum force of static friction divided by the normal force must stay below 0.30 for the block to remain there.

But how can i find the acceleration? i have nothing to work with :confused:
 
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There're two forces here:
-the gravitational force pulling the block down
-fricton in the opposite direction
the gravitational force is mg and the friction is N(normal) * static coefficient.
You know that the normal force comes from the acceleration of the cart (2 law). In order for the block to fall the fricton has to be at least equal to mg. After finding what the friction has to be, from that find the normal and then the acceleration.
 
You might go back and check the wording of the question. The "minimum acceleration" for the block to stay in place would be 0. If the cart doesn't accelerate, I guarantee the block doesn't move, and it's hard to get less than 0. Something's amiss here.
 
I think that the question asks for the minimum acceleration of the cart which causes friction on the block.
 
I would guess that the problem actually asked for the maximum acceleration at which the block does not move. Of course that is the minimum acceleration at which there is (sliding) friction on the block.

frigid: You know the mass of the block so you can calculate its weight- the force downward on the cart. You know the coefficient of static friction so you can calculate the force at which the block will start to move. You know the mass of the cart (and block) so you can calculate the acceleration on the cart corresponding to that force.
 
Sorry guys, i failed to mention: the front of the cart is flat and vertical, it's not a slope like I'm guessing you assumed it was.

So i find the weight of the block: 0.2*9.8=1.96N
Fn = 1.96*0.3 = 0.59N

How do I find the acceleration? I don't have a velocity :rolleyes:
 
I assume that the block is literally on the front face of the cart. (Some posters think it's on top of the cart, towards the front.)

In order for the block not to slide down, what must the friction force be? In order to get that amount of friction, what must the normal force be? Given that normal force, what must the acceleration be?

Please solve things symbolically--don't plug in numbers until the last step.
 
Ff: 0.2*9.8 = 1.96N
Fn: 1.96 / 0.3 = 6.53N
Fa - Ff = ma
6.53 - 1.96 = 0.1*a
4.57 = 0.1*a
a = 45.7m/s^2

The concern i have is that i didn't use the kart's 10kg no where in my calculations. Is this kosher?
 
frigid said:
Ff: 0.2*9.8 = 1.96N
Correct.
Fn: 1.96 / 0.3 = 6.53N
Correct.
Fa - Ff = ma
6.53 - 1.96 = 0.1*a
Incorrect. Why are you subtracting the friction force? Vertical forces are in equilibrium; the only forces you need consider are horizontal forces. What is "0.1"?

As I suggested, please solve it symbolically before plugging in numbers.
The concern i have is that i didn't use the kart's 10kg no where in my calculations. Is this kosher?
Fine by me.
 

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