Academic Physics
Forces in Equilibrium
07.3HW
1.
A force F = (17N, 68o) is applied to a book sitting on a table. What are polar coordinates of the force required to be applied
so that the book maintains its equilibrium?
The force would need to be exactly the opposite of what is being applied; in other words, the force would
have to be at 180o from where it is applied. The coordinates are (17N, 248 o).
2.
A force F = (-5, 10)kN is applied to a chair in static equilibrium. What are the polar coordinates of the force required to be
applied to the chair to keep it in equilibrium?
5 + 10 = R
The resultant force is at:
R = 11.18 kN
180 - 63.43 = 116.57
tan  = (10/5)
The force to keep equilibrium would be the opposite of this force:
 = 63.43
(11.18 kN, 296.57 )
2
3.
2
2
sin 27 = y/15N
y = 6.81N
cos 27 = x/15N
x = 13.37N
o
27
The force to keep equilibrium would be the
opposite of this force:
(-13.37, -6.81)N
A 12-kg box is pulled along the floor with a force of 85N. The coefficient of friction, , between the floor and the box is
0.45. Calculate the force required for the box to maintain its state of dynamic equilibrium.
Fn = 120N
Ff = (0.45)(120N) = 54N
Fequilibrium = 85N – 54N = 31N
5.
o
A force F = (15N, 27o) is applied to a crate. What are the Cartesian coordinates of the force required to be applied to the
crate to have it maintain its static equilibrium?
15N
4.
o
so the equilibrium force would be -31N
(Over>>>>)
Three forces are applied to a shipping crate. Find the x- and y- components of the force required to counteract the forces F1
= (100N, 315o), F2 = (85N, 210o), and F3 = (60N, 150o) so the crate remains in equilibrium.
sin 315 = y / 100N
y = -70.71N
sin 210 = y / 85N
y = -42.50N
sin 150 = y / 60N
y = 30N
cos 315 = x / 100N
x = +70.71N
cos 210 = x / 85N
x = -73.61N
cos 150 = x / 60N
x = -51.96N
(70.71, -70.71)N + (-73.61, -42.50)N + (-51.96, 30)N = (-54.86, -83.21)N
Academic Physics
Forces in Equilibrium
07.3HW
so the equilibrium force would be the opposite of the resultant: (+54.86, +83.21)N
6.
In order to keep the ball in static equilibrium, what are the polar coordinates of
the force that must be applied to counteract the four forces shown?
See the diagram at the right. Please note, the diagram is not drawn to scale.
Sin 15 = y / 18N
Y = 4.66N
Cos 15 = x / 18N
X = 17.39N
sin 117 = y / 24N
y = 21.38N
cos 117 = x / 24N
x = -10.90N
F2 = 24N
27o
8o
F1 = 18N
15o
22o
F3 = 11N
F4 = 14N
Sin 188 = y / 11N
Y = -1.53N
Cos 188 = x / 11N
X = -10.89N
sin 338 = y / 14N
y = -5.24N
cos 338 = x / 14N
x = 12.98N
(17.39, 4.66)N + (-10.90, 21.38)N + (-10.89, -1.53)N + (12.98, -5.24)N = (8.58, 19.27)N = resultant
so the equilibrium force would be the opposite of the resultant = Fequilibrium = (-8.58, -19.27)
Feq2 = 8.582 + 19.272
Feq = 21.09N
Tan  = y / x
tan  = 19.27 / 8.58
Force which keeps equilibrium = (21.09N, 267.94 o)
 = 87.94 + 180 = 267.94o