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16 SLOT ATTACHED TOP WOOD JEWELRY DISPLAY CASE BOX US $11.69
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36 RING ATTACHED TOP WOOD JEWELRY DISPLAY CASE US $17.99
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Vintage Baribocraft Wood Nutcracker Bowl Nut Cracker Attached to Top US $38.00
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OTHER RELATED PRODUCTS:
![]() |
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16 SLOT ATTACHED TOP WOOD JEWELRY DISPLAY CASE BOX US $11.69
|
36 RING ATTACHED TOP WOOD JEWELRY DISPLAY CASE US $17.99
|
Vintage Baribocraft Wood Nutcracker Bowl Nut Cracker Attached to Top US $38.00
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Can someone help me with this physics problem?
A rope attached to a 16.0 kg wood sled pulls the sled up a 19.0' snow-covered hill. A 6.00 kg wood box rides on top of the sled. If the tension in the rope steadily increases, at what value of the tension does the box slip?
The increasing tension in the pulling rope means that the sled is accelerating up the slope. The box sitting on the sled must also accelerate. The only force which can accelerate the box is friction between the box and the sled. That friction ( static friction ) will have some maximum value and once the acceleration required cannot be supplied by the friction the box will slip.
So, you need to know what the coefficient of static friction (µ) is between the box and the sled.
T ( tension ) = 22 sin19˚ (weight component down slope ) + 22 .a ( accelerating force )
On the box Friction = 6 . a = N . µ = 6 . cos 19˚ . µ
a = µ cos19˚
T = 22 sin 19˚ + 22 . µ . cos 19˚
Looks good - just needs a value for "µ"
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