A block of mass 5 kg rides on top of a second block of mass 10 kg. A person attaches a string to the bottom block and pulls the system horizontally across a frictionless surface. Friction between the two blocks keeps the 5 kg block from slipping off. If the coefficient of static friction is 0.350, what maximum force can be exerted by the string on the 10.0 kg block without causing the 5.00 block to slip?


Top Block: mg downwards, normal force up, fs to right.
Bottom Block : mg1 down, mg2 down, normal force up, tension right.

My friend claims that there should also be an fs acting on the Bottom block to the left, but I cant seem to see why. Could someone explain it to me?


2 Answers 2


Your friend is right. There is an equal but opposite friction force acting leftward on the bottom block.

While the top block is being pulled along by a friction force, the bottom block is being "held back" the same amount. If the top block wasn't there, the bottom block would be able to accelerate faster. But since it has to bring the top block along, it is slowed down - the friction is what makes them stick together.

This "holding back" force is equal but opposite static friction, which is a result of Newton's 3rd law.


Newton's third law.

  • fs to right is the frictional force on the top block due to the bottom block

  • fs to left is the frictional force on the bottom block due to the top block.

The horizontal frictional forces are the means by which the two blocks "communicate" with one another and try to prevent relative movement between the two blocks.

  • fs to the right on the top block is trying to increase the velocity of the top block so that it is moving at the same velocity as the bottom block

  • fs to the left on the bottom block is trying to decrease the velocity of the bottom block so that It is moving at the same velocity as the top block.


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