If an automobile of mass m rounds a curve, then its inward vector component of acceleration a N N is caused by the frictional force F of the road. Thus, it follows from the vector form of Newton’s second law [Equation (19)] that the frictional force and the normal component of acceleration are related by the equation F = m a N N . Thus, F = m k d s d t 2 Use this result to find the magnitude of the frictional force in newtons exerted by road on a 500 kg go-cart driven at a speed of 10km/h around a circular track or radius 15m. [ N o t e : 1 N = 1 kg . m / s 2 . ]
If an automobile of mass m rounds a curve, then its inward vector component of acceleration a N N is caused by the frictional force F of the road. Thus, it follows from the vector form of Newton’s second law [Equation (19)] that the frictional force and the normal component of acceleration are related by the equation F = m a N N . Thus, F = m k d s d t 2 Use this result to find the magnitude of the frictional force in newtons exerted by road on a 500 kg go-cart driven at a speed of 10km/h around a circular track or radius 15m. [ N o t e : 1 N = 1 kg . m / s 2 . ]
If an automobile of mass m rounds a curve, then its inward vector component of acceleration
a
N
N
is caused by the frictional force F of the road. Thus, it follows from the vector form of Newton’s second law [Equation (19)] that the frictional force and the normal component of acceleration are related by the equation
F
=
m
a
N
N
.
Thus,
F
=
m
k
d
s
d
t
2
Use this result to find the magnitude of the frictional force in newtons exerted by road on a 500 kg go-cart driven at a speed of 10km/h around a circular track or radius 15m.
[
N
o
t
e
:
1
N
=
1
kg
.
m
/
s
2
.
]
Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.
A seasoned parachutist went for a skydiving trip where he performed freefall before deploying
the parachute. According to Newton's Second Law of Motion, there are two forcës acting on
the body of the parachutist, the forces of gravity (F,) and drag force due to air resistance (Fa)
as shown in Figure 1.
Fa = -cv
ITM EUTM FUTM
* UTM TM
Fg= -mg
x(t)
UTM UT
UTM /IM LTM
UTM UTM TUIM
UTM F UT
GROUND
Figure 1: Force acting on body of free-fall
where x(t) is the position of the parachutist from the ground at given time, t is the time of fall
calculated from the start of jump, m is the parachutist's mass, g is the gravitational acceleration,
v is the velocity of the fall and c is the drag coefficient. The equation for the velocity and the
position is given by the equations below:
EUTM PUT
v(t) =
mg
-et/m – 1)
(Eq. 1.1)
x(t) = x(0) –
Where x(0) = 3200 m, m = 79.8 kg, g = 9.81m/s² and c = 6.6 kg/s. It was established that the
critical position to deploy the parachutes is at 762 m from the ground…
Write Newton’s Second Law of Motion for three-dimensionalmotion with only the gravitational force (acting in the z-direction).
An ant walks due east at a constant speed of 2 mi/hr on a sheet of paper that rests on a table. Suddenly,
the sheet of paper starts moving due southeast at v2 mi/hr. The following is a correct solution for finding
the resultant speed and direction of the ant relative to the table.
Set up a coordinate system so that the positive y-axis corresponds to north and the positive x-axis corresponds
to east. Then the velocity vector for the ant relative to the paper is given by (2,0). The velocity vector for
the paper relative to the table is given by (v2 cos (-4), v2 sin (-5))= (1, –1).
Therefore, the resultant velocity vector is given by (2,0) + (1, –1) = (3, –1). The speed of the ant relative
to the table is v32 + 1² = v10 mi/hr and the direction the ant is traveling relative to the table is found by
(금)
using tan 0 =
-1
where 0 represents the angle made with the positive r-axis. In this case 0 = arctan
3
3
Chapter 12 Solutions
Calculus Early Transcendentals, Binder Ready Version
Precalculus: Mathematics for Calculus (Standalone Book)
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