Question

As shown in the figure below, two forces are applied horizontally to two blocks in contact. The horizontal surface on which the blocks slide frictionless. if F = 6.0N and M=1.0kg, what is the magnitude of the force exerted on the large block of mass 3M by the small block of mass 2M.

https://i.gyazo.com/25f5ef792c93e6b6a36a11cfb984d5a5.png

Answer 1

Let's first find the acceleration of the system of two blocks from the Newton's Second Law of Motion:

$$2F-F=(3M+2M)a,$$ $$a=\dfrac{F}{5M}=\dfrac{6.0\ N}{5\cdot1.0\ kg}=1.2\ \dfrac{m}{s^2}.$$

Then, we can find the net force on the first block with mass $3M$:

$$F_{1,net}=3Ma=3\cdot1.0\ kg\cdot1.2\ \dfrac{m}{s^2}=3.6\ N.$$

Finally, we can find the magnitude of the force exerted on the large block of mass 3M by the small block of mass 2M:

$$2F-F_{21}=F_{1,net},$$ $$F_{21}=2F-F_{1net}=2\cdot6.0\ N-3.6\ N=8.4\ N.$$

Answer:

$F_{21}=8.4\ N.$