# b) A horizontal rope pulls a 10kg wood sled across frictionless snow.A 5kg wood box rides on top of the sled. What is the largest tension force for which the box doesnt slip.

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### 1 Answer

You should select the coefficient of friction mu between two pieces of wood of 0.2 (wet wood) or 0.5 (clean wood).

Considering `mu = 0.5` and using the equation that relates the friction force, mu and normal force yields:

`{(F_f = mu*N),(N = m*g):} => F_f = mu*m*g`

Hence, using `mu = 0.5, m = 5Kg` and `g = 9.8 m/s^2` yields:

`F_f = 0.5*5*9.8 = 24.5 N`

Using the Newton's second law yields:

`F_f = m*a => 24.5 = 5*a => a = 4.9 m/s^2`

Since you know the acceleration, you may evaluate the largest tension in the rope such that:

`T = (M + m)*a`

M represents 10 Kg wood piece

m represents 5Kg wood box

`T = (10+5)*4.9 => T = 73.5N`

**Hence, evaluating the largest tension in rope for which the `5Kg` box does not slip yields `T = 73.5N.` **