The forces F 1 = 6 i − j , F 2 = − 7 i − 2 j and F 3 = 9 i + 3 j act on an object. What additional force F is needed for the object to be in static equilibrium?
The forces F 1 = 6 i − j , F 2 = − 7 i − 2 j and F 3 = 9 i + 3 j act on an object. What additional force F is needed for the object to be in static equilibrium?
Solution Summary: The author calculates the additional forces F needed for the object to be in static equilibrium. The resultant force is 8i.
The forces
F
1
=
6
i
−
j
,
F
2
=
−
7
i
−
2
j
and
F
3
=
9
i
+
3
j
act on an object. What additional force F is needed for the object to be in static equilibrium?
Three forces act on an object: F = (-8,–5), F, = (0,1), F; = (4, – 7). Find the net
%3D
force acting on the object.
Brazilian soccer star Marta has a penalty kick in the quarter-final match. She kicks the soccer ball from ground level with
the (x, y)-coordinates (81, 19) on the soccer field shown in the figure and with initial velocity Vo = 8i – 4j + 31k ft/s.
Assume an acceleration of 32 ft/s? due to gravity and that the goal net has a height of 8 ft and a total width of 24 ft.
105 ft
12
12
105 ft
165 ft
165 ft
Determine the position function that gives the position of the ball t seconds after it is hit.
(Use symbolic notation and fractions where needed.)
r(t) = r;(1)i + r,(f)j+ r¢(1)k
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…
Thomas' Calculus: Early Transcendentals (14th Edition)
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Area Between The Curve Problem No 1 - Applications Of Definite Integration - Diploma Maths II; Author: Ekeeda;https://www.youtube.com/watch?v=q3ZU0GnGaxA;License: Standard YouTube License, CC-BY