Consider the horse as the 'system', then there are three forces acting on the horse
(i) Downward gravitational force (m_hg)
(ii) Force exerted by the road (F_r)
(iii) Backward force exerted by the cart (F_c)
It is shown in the following figure.
F_r - Force exerted by the road on the horse
F_c - force exerted by the cart on the horse
F_r^丄 - Perpendicular component of F_r = N
F_r^||-Parallel component of F, which is reason for forward movement.

The force exerted by .the road can be resolved into parallel and perpendicular components, The perpendicular component balances the downward gravitational force. There is parallel component along the forward direction. It is greater than the backward force (F_c). So there is net force along the forward direction which causes the forward movement of the horse.
If we take the cart as the system, then there are three forces acting on the cart.
(i) Downward gravitational force (m_cg)
(ii) Force exerted by the road (F_r')
(iii) Force exerted by the horse (F_h)

It is shown in the figure
The force exerted by the road (\(\vec { { F }_{ r } } \)) can be resolved into parallel and perpendicular components. The perpendicular component cancels the downward gravity (m_cg)
Parallel component acts backwards and the force exerted by the horse (\(\vec { { F }_{ h } } \)) acts forward. Force (\(\vec { { F }_{ h } } \)) is greater than the parallel component acting in the opposite direction. So there is an overall unbalanced force in the forward direction which causes the cart to accelerate forward.
If we take the cart + horse as a system, then there are two forces acting on the system.
(i) Downward gravitational force (m_h + m_c)g
(ii) The force exerted by the road (F_r) on the system.
It is shown in the following figure.

(iii) In this case the force exerted by the road (F_r) on the system (cart + horse) is resolved in to parallel and perpendicular components. The perpendicular component is the normal force which cancels the downward gravitational force (m_h +m_c)g. The parallel component of the force is not balanced, hence the system (cart + horse) accelerates and moves forward due to this force.

(i) Choose a suitable inertial coordinate system to analyse the problem. For most of the cases we can take Earth as an inertial coordinate system.
(ii) Identify the system to which Newton's laws need to be applied. The system can be a single object or more than one object.
(iii) Draw the free body diagram.
(iv) Once the forces acting on the system are identified, and the free body diagram is drawn, apply Newton's second law. In the left hand side of the equation, write the forces acting on the .system in vector notation and equate it to the right hand side of equation which is the product of mass .and acceleration. Here, acceleration should also be in vector notation.
(v) If acceleration is given, the force can be calculated. If the force is given, acceleration can be calculated.
By following the above steps: We fix the inertial coordinate system on the. ground as shown in the figure.

The forces acting on the mango are
(i) Gravitational force exerted by the Earth on the mango acting downward along negative y-axis.
(ii) Tension (in the cord attached to the mango) acts upward along positive y-axis.
The free body diagram for the mango is shown in the figure.
\(\vec { { F }_{ g } } =mg(-\hat { j } )=-mg\hat { j } \)
Here, mg is the magnitude of the gravitational force and \((-\hat { j } )\) represents the unit vector in negative y direction.
\(\vec { T } =T\hat { j } \)



Here T is the magnitude of the tension force and \((-\hat { j } )\) represents the unit vector in positive y direction.
\({ \vec { F } }_{ net }={ \vec { F } }_{ s }+{ \vec { T } }_{ g }=-mg\hat { j } +T\hat { j } =(T-mg)\hat { j } \)
From Newton's second law \({ \vec { F } }_{ net }=m\vec { a } \)
Since the mango is at rest with respect to us (inertial coordinate system) the acceleration is zero (\(\vec { a } =0\))
So \({ \vec { F } }_{ net }=m\vec { a } =0\)
\((T-mg)\hat { j } =0\)
By comparing the components on both sides of the above equation, we get T - mg = 0
So the tension force acting on the mango is given by T - mg
Mass of the mango m = 400g and g = 9.8 ms^-2 Tension acting on the mango is T = 0.4\(\times\)9.8 = 3.92 N.

(i) If the object is at rest and no external force is applied on the object, the static friction acting on the object is zero (f_s = 0).
(ii) If the object is at rest, and there is an external force applied parallel to the surface, then the force of static friction acting on the object is exactly equal to the external force applied on the object (f_s = F_ext). But still the static friction Is is less than μ_sN.
(iii) When object begins to slide, the static friction (f_s) acting on the object attains maximum.

If a force (F) acts on the object in a very short interval of time (M), from Newton's second law in magnitude form
Fdt = dp
Integrating over time from an initial time t_i to a final time t_f, we get
\(\int _{ i }^{ f }{ dp } =\int _{ { t }_{ i } }^{ { t }_{ f } }{ Fdt } \)
pf-pi = \(\int _{ { t }_{ i } }^{ { t }_{ f } }{ Fdt } \)
pi = initial momentum of the object at time t_i
P_t = final momentum of the object at time t_f.
pf - pi = Δp change in momentum of the object during the time interval
tf - ti = Δt
The integral \(\int _{ { t }_{ i } }^{ { t }_{ f } }{ Fdt } \)=J is called the impulse and it is equal to change in momentum of the object.
If the force is constant over the time interval, then
\(\int _{ { t }_{ i } }^{ { t }_{ f } }{ F } dt=\int _{ i }^{ f }{ dp } \) = F(t_f - t_i) = FΔt
FΔt = Δp