- #1

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Air resistance/Drag is neglected. Wouldn't the box remain on the sled indefinetely as long as it didn't accelerate to fast? If not then I would use the coefficient of static friction right?

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- Thread starter Comtrend
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- #1

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Air resistance/Drag is neglected. Wouldn't the box remain on the sled indefinetely as long as it didn't accelerate to fast? If not then I would use the coefficient of static friction right?

- #2

Andrew Mason

Science Advisor

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You would need to know the coefficient of static friction between the box and sled to find the maximum static friction force. That is the maximum force that the sled can apply to the box. If the tension exceeds that force, the box will slide off.Comtrend said:I have a question where a horizontal rope pulls on a 10kg wood sled over a frictionless surface. On that sled is a wood box. It is asking what the largest tesion force in the rope would be where the box doesn't slip off.

Air resistance/Drag is neglected. Wouldn't the box remain on the sled indefinetely as long as it didn't accelerate to fast? If not then I would use the coefficient of static friction right?

AM

- #3

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for the box:

[tex]F_f=m_{box} a[/tex]

and for the sled

[tex]F-F_f=m_{sled} a[/tex]

As we can see, we have 3 unknows and only 2 eqs. So you must provide some information about the friction at the contact surface.

- #4

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So then Tension equals 14.72N?

- #5

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I don't know how to get mew so it is just an m.

- #6

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You have to find F from the eq. below:

[tex]\frac{F-F_f}{F_f}=\frac{m_{sled}}{m_{box}}[/tex]

and you'll obtain the tension in the rope.

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