A secret agent pushes two mysterious crates across a frozen river in the dark of night. The crates slide friction-lessly. Their masses are 150 kg (A) and 50 kg (B). The agent's hands exert a 100 N force on A (see Fig. 5-19). How much force does A exert on B? If there is no friction, can either block's acceleration be zero? Therefore, can `sum vec(F)_("on A")=0`? Therefore, can the force exerted on A by B (which is equal in magnitude to that exerted on B by A) "balance" (be equal and opposite to) the force exerted by the agent?

Explanation

With no friction, the unopposed force exerted by the agent makes both crates accelerate. If crate A is accelerating, the forces on A cannot be equal and opposite.
Calculations: (a) To find a force on block B, we apply Newton.s second law to B. Using a free-body diagram of B (left), we get
`sun F_("x on B")=F_("on B by A")=m_(B)alpha_("Bx")=m_(B)alpha_(B)." "(5-21)`
This leaves us with two unknowns, `F_("on B by A")` and `alpha_(B)`.
(b) However, if we apply the second law to block A as well (see free-body diagram of A at left), we get
`sun F_("x on A)=F_("on A by agent")-F_("on A by B")=m_(A)alpha_(Ax)=m_(A)alpha_(A)." "(5-22)`
However, we know that `F_("on A by B")=F_("on B by A")`. Also, `alpha_(A)=alpha_(B)` (label this common acceleration a). Thus, from Eqs. (5-21) and (5-22), we get
`F_("on B by A")=m_(B)alpha" "(5-23)`
and `" "F_("on A by agent")-F_("on B by A")=m_(A)alpha," "(5-24)`
which we can solve for the two unknowns `F_("on B by A")` and a. We use Eq. (5-23) to substitute for `F_("on B by A")` in Eq. (5-24):
`F_("on A by agent")-m_(B)alpha=m_(A)alpha`.
Then
`F_("on A by agent")=m_(B)alpha+m_(A)alpha=(m_(B)+m_(A))alpha`
and
`alpha=(F_("on A by agent"))/(m_(B)+m_(A))=(100" N")/(50" kg"+150" kg")=0.50" m"//"s"^(2)" "(5-25)`
Substituting this value and `m_(A)=50" kg"` into Eq. (5-23) gives us
`F_("on B by A")=(50" kg")(0.50" m"//"s"^(2))=25" N"`.
We have found that `F_("on B by A")` is one quarter of the force exerted by the hand on A. Likewise, `m_(B)` is one quarter of the combined mass `m_(A)+m_(B)` pushed by the hand. As you sould expect from Newton.s second law of motion, to impart the same acceleration to a mass one-fourth as great requires one fourth the force.