Penguin forces problem.?

Question

A 3.8 kg penguin sits on a 10 kg sled. A horizontal force of 45 N is applied to the sled, but the penguin attempts to impede the motion by holding onto a cord attached to a wall. The coefficient of kinetic friction between the sled and snow, as well as that between the sled and the penguin, is 0.20.

I drew and FBD but I don't understand how to find the tension of the cord. I know I multiply .20 with something but I dont know what. Thanks.

Answer 1

There are two FBDs: One for the penguin and one for the sled.

Sled:

Positive in the direction of the 45 N force
The normal force of the sled + the mass of the penguin results in the friction of the sled to the snow:

-13.8*9.81*.2
-27.1

Assume the penguin is sliding on the sled
There is a force due to friction of the penguin
-3.8*9.81*.2
-7.46

This was all a waste of time so far since the penguin is not moving with respect to the cord, the sled is sliding from underneath it. Therefore there is no acceleration of the penguin.
The tension in the cord is equal to the frictional force between the penguin and the sled:
-7.46 N
(note the negative based on the convention I started with. This force opposes the 45N force)

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