91Ó°ÊÓ

The given $FIGUREP7.51$follows the loaded truck with the crate in which the crate moves forward when the brake force is applied to the truck. the crate interacting twice with the truck—a friction force parallel to the surface of the truck bed and a normal force perpendicular to this surface. The truck interacts similarly with the road surface, but notice that the crate does not interact with the ground; there’s no contact between them.

The two interactions within the system are each an action/reaction pair, so this is a total of four forces. You can also see four external forces crossing the system boundary, so the free-body diagrams should show a total of eight forces.

Now we are ready to write the newton's second law for Crate

$$\begin{gathered} \underset{i}{∑}{\left(F_{oncrate}\right)}_{i,x}=7.0N=m_c{a}_c \\ \underset{i}{∑}{\left(F_{oncrate}\right)}_{i,y}=9.8N-9.8N=0 \end{gathered}$$

For the Truck

$$\begin{gathered} \underset{i}{∑}{\left(F_{ontruck}\right)}_{i,x}=21.0N-7.0N=13N=m_T{a}_T \\ \underset{i}{∑}{\left(F_{ontruck}\right)}_{i,y}=29.4N-19.6N-9.8N=0 \end{gathered}$$

Therefore, the net force acting on the truck when crate sliding down during braking is localid="1649776600779" $13.0N$, The net force acting on the crate sliding down during braking is$7.0N$