All Solutions
Page 160: Lesson Check
$$
begin{gather*}
sum vec{textbf{F}} = mvec{textbf{a}}
end{gather*}
$$
An object moving from rest or changing its speed or direction is said to have a nonzero acceleration.
Thus, according to $textbf{Newton’s second law}$, for an object to have a nonzero acceleration, there must be $textbf{a nonzero net force acting on it}$.
This net force can be a single force or multiple forces where their vector sum has a nonzero magnitude.
In this situation, the man exerts a force on the wall but the wall stays stationary because there is another force (friction force) exerted on the wall opposite in direction to that exerted by the man.
– There is an action-reaction pair, in this case, the action force is the force exerted by the man on the wall and the reaction force is the force exerted by the wall on the man.
According to $textbf{Newton’s third law}$, both forces (action and reaction) have the same magnitude and opposite in direction.
– The action force is the force exerted by the man on the wall and the reaction force is the force exerted by the wall on the man. And they are equal in magnitude and opposite in direction.
(b) There is a nonzero force acting on it.
(c) According to $textbf{Newton’s second law}$
– For example, a body thrown up in the air while it is still going up, and a body moving in a circular path.
– For every action force acting on an object, there is a reaction force acting on a different object.
– The action and reaction forces are equal in magnitude and opposite in direction.
Thus, the force exerted by the car on the ground is $vec{textbf{F}}_1$ (action).
According to $textbf{Newton’s third law}$, the ground pushes back on the car (reaction), so that the force exerted by the ground on the car is $vec{textbf{F}}_2$.
(b) According to $textbf{Newton’s third law}$, both forces $vec{textbf{F}}_1$ and $vec{textbf{F}}_2$ are equal in magnitude and opposite in direction.
Thus, the right choice is (2) $vec{textbf{F}}_2 = – vec{textbf{F}}_1$.
(b) The right choice is (2) $vec{textbf{F}}_2 = – vec{textbf{F}}_1$
From $textbf{Newton’s second law}$ for single force, the force $F$ exerted on an object of mass $m$ moving with acceleration $a$ is given by:
$$
begin{gather*}
F = matag{1}
end{gather*}
$$
The mass of the baseball is $m = 0.15$ kg and its acceleration is $a = 12text{ m/s}^2$.
We simply plug our values for $m$ and $a$ into equation (1), so we get the force required:
$$
begin{gather*}
F = (0.15text{ kg})cdot(12.0text{ m/s}^2) = 1.80text{ N}\\
therefore quad large boxed{F = 1.80text{ N}}
end{gather*}
$$
F = 1.80text{ N}
$$
From $textbf{Newton’s second law}$ for multiple forces, the net force acting on an object is equal to the mass of the object times its acceleration:
$$
begin{gather*}
sum vec{textbf{F}} = mvec{textbf{a}}tag{1}
end{gather*}
$$
The mass of the airplane is $m = 3.50times 10^{5}$ kg and the net braking force acting on it is $sum vec{textbf{F}} = -4.30times 10^{5}text{ N}$ (the negative is because the braking force is always opposite to the motion).
We simply plug our values for $m$ and $sum vec{textbf{F}}$ into equation (1) and evaluate for $vec{textbf{a}}$, so we get the acceleration of the airplane:
$$
begin{gather*}
vec{textbf{a}} = dfrac{sum vec{textbf{F}}}{m} = dfrac{-4.30times 10^{5}text{ N}}{3.50times 10^{5}text{ kg}} = -1.23text{ m/s}^2\\
therefore quad large boxed{vec{textbf{a}} = -1.23text{ m/s}^2}
end{gather*}
$$
vec{textbf{a}} = -1.23text{ m/s}^2
$$
From **Newton’s second law** for multiple forces, the net force acting on an object is equal to the mass of the object times its acceleration:
$$begin{gather*}
sum vec{textbf{F}} = mvec{textbf{a}}tag{1}
end{gather*}$$
From **Newton’s third law**,
– For every action force acting on an object, there is a reaction force acting on a different object.
– The action and reaction forces are equal in magnitude and opposite in direction.
The mass of the car is $m_c = 1400$ kg, the mass of the trailer is $m_t = 560$ kg and the their acceleration is $vec{textbf{a}} = 1.85text{ m/s}^2$.
(a) The net force exerted on the trailer is the force exerted by the car.
So, we plug our values for $m_t$ and $vec{textbf{a}}$ into equation (1), so we get the net force exerted by the car on the trailer:
$$begin{gather*}
sum vec{textbf{F}}_t = (560text{ kg}) cdot (1.85text{ m/s}^2) = \\
therefore quad large boxed{sum vec{textbf{F}}_t = 1036text{ N}}
end{gather*}$$
$$begin{gather*}
vec{textbf{F}}_{tc} = -sum vec{textbf{F}}_t\\
therefore quad large boxed{vec{textbf{F}}_{tc} = -1036text{ N}}
end{gather*}$$
$$begin{gather*}
sum vec{textbf{F}}_c = (1400text{ kg})cdot (1.85text{ m/s}^2) = 2590text{ N}\\
therefore quad large boxed{sum vec{textbf{F}}_c = 2590text{ N}}
end{gather*}$$
**Note:** The net force exerted on the car is the sum of the forward force exerted by the motor and the backward pulling force exerted by the trailer.
Where the total force exerted by the motor of the car is:
$$begin{gather*}
vec{textbf{F}}_{tot} = (1400text{ kg} + 560text{ kg}) cdot (1.85text{ m/s}^2) = 3626 text{ N}
end{gather*}$$
(b) $vec{textbf{F}}_{tc} = -1036text{ N}$
(c) $sum vec{textbf{F}}_c = 2590text{ N}$
