Assuming that force is in newtons and distance is in meters, find the work done by the resultant of the constant forces F 1 = i − 3 j + k and F 2 = i − 2 j + 2 k acting on a particle that moves on a straight line from P − 1 , − 2 , 3 to Q 0 , 2 , 0 .
Assuming that force is in newtons and distance is in meters, find the work done by the resultant of the constant forces F 1 = i − 3 j + k and F 2 = i − 2 j + 2 k acting on a particle that moves on a straight line from P − 1 , − 2 , 3 to Q 0 , 2 , 0 .
Assuming that force is in newtons and distance is in meters, find the work done by the resultant of the constant forces
F
1
=
i
−
3
j
+
k
and
F
2
=
i
−
2
j
+
2
k
acting on a particle that moves on a straight line from
P
−
1
,
−
2
,
3
to
Q
0
,
2
,
0
.
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…
A 50-kg wagon is pulled to the right by a force F1 making an angleof 30◦ with the ground. At the same time, the wagon is pulled to theleft by a horizontal force F2.(a) Find the magnitude of F1 in terms of the magnitude of F2 if thewagon does not move.
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
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