: SECOND CONDITION OF EQUILIBRIUM :
: TORQUE : Γ :
- The tendency of a body to rotate.
F2
ℓ2 ℓ1
O
ℓ3
- The product of force and lever arm.
Lever arm - the perpendicular
distance of the point of rotation to
the line of action of the force.
F3
MKS: N•m
ΓOF2 = + F2ℓ2
CGS: Dyne•cm
ΓOF3 = + F3ℓ3
FPS: lb•ft
ΓOF4 = F4ℓ4 = 0
: SIGN CONVENTION :
Counter Clockwise : + Γ
F4
ΓOF1 = - F1ℓ1
: UNITS OF TORQUE :
Clockwise : - Γ
F1
If the line of force passes through the
axis of rotation, the torque is zero.
Now:
ƩΓaxis = 0
Problem 8: A 71.0-kg boulder is placed on a 2.00-m long board, at a point that is 1.60
m form one end. Cliff and Will support the board at each end so that it is horizontal.
Cliff is nearest the boulder. If the weight of the board is negligible, what force does
each apply to the board?
WB
Solution:
WB = (71)(9.8)
WB = 695.80 N
R1
R2
1.60 m
0.40 m
Note:
ƩΓ1 = 0
ƩΓ2 = 0
ƩFy = 0
Γ1R2 - Γ1WB = 0
- Γ2R1 + Γ2WB = 0
R1 + R2 - WB = 0
R2(2.0) = 695.80(1.60)
R1(2.0) = 695.80(0.40)
R2 = 556.64 N
R1 = 139.16 N
(Force of Cliff)
(Force of Will)
R1 = - R2 + WB
R1 = - 556.64 + 695.80
R1 = 139.16 N
Problem 9: The mass of an automobile is 1160 kg, and the horizontal distance
between its front and rear axles is 2.54 m. The center of gravity of the car is between
the front and rear tires, and the horizontal distance between the center of gravity and
the front axle is 1.02 m. Determine the normal force that the ground applies to each of
the two front wheels and to each of the two rear wheels.
Solution:
ƩΓF = 0
W
2NR(2.54) - W(1.02) = 0
2NR(2.54) = 11,368(1.02)
2NF
W = 1160(9.8)
W = 11,368 N
NR =
2NR
11,368(1.02)
2(2.54)
NR = 2,282.55 N
W = 11,368 N
ƩFY = 0
2NF + 2NR - W = 0
2NF = 11,368 - 2(2,282.55)
2NF
1.02 m
1.52 m
2NR
NF = 3,401.44 N
Problem: Two people are carrying a uniform horizontal ladder that is 6.00 m long
and weighs 400 N. If one person applies an upward force equal to 180 N at one
end, at what point does the other person lift?
Solution:
3.0 m
3.0 m
WL = 400 N
∑FY = 0
RA + RB - WL = 0
RB = WL - RA
RA = 180 N
X=?
RB = ?
RB = 400 - 180
RB = 220 N
∑ΓA = 0
RB(X) - WL(3.0) = 0
220(X) = 400(3.0)
X = 5.45 m
Problem: A diving board 3.00 m long is supported at a point 1.00 m from the
end, and a diver weighing 580 N stands at the free end. The diving board is on
uniform cross section and weighs 320 N. Find (a) the force at the support point;
(b) the force at the end that is held down.
ƩΓB = 0
2.0 m
RA(1.0) - WB(1.5) - WD(3.0) = 0
1.0 m
RA(1.0) = (320)(1.5) + (580)(3.0)
Consider the FBD:
WB 1.50 m
RB 1.50 m
RA = 2,220 N
WD
Alternative Solution for RA:
∑FY = 0
1.00 m
RA
RA - RB - WB - WD = 0
ƩΓA = 0
RA = RB + WB + WD
RB(1.0) - WB(0.5) - WD(2.0) = 0
RA = 1,320 + 320 + 580
RB(1.0) = (320)(0.5) + (580)(2.0)
RB = 1,320 N
RA = 2,220 N
Problem: Two people carry a heavy electric motor by placing it on a light board 2.00 m
in length. One person lifts at one end with a force of 700 N, and the other lifts the
opposite end with a force of 500 N. What is the weight of the motor, and where along
the board is its center of gravity located?
Solution:
(2 - X)
X
∑Γ1 = 0
WM = ?
R2(2.0) - WM(X) = 0
- 1,200(X) = - 500(2.0)
2.0 m
R1 = 700 N
∑FY = 0
R1 + R2 - WM = 0
R1 + R2 = WM
WM = 700 + 500
WM = 1,200 N
R2 = 500 N
X = 0.83 m
(From the 700 N support.)
Note:
(2 - X) = 2 - 0.83
(2 - X) = 1.17 m
(From the 500 N support.)
: CENTER OF GRAVITY OR CENTER OF MASS :
It is a point where the weight is acting.
Consider:
Horizontal Location:
WT(XT) = W1(X1) + W2(X2) + W3(X3)
mT
or
c.g.
XT
mT(XT) = m1(X1) + m2(X2) + m3(X3)
m1
m2
Vertical Location:
c.g.1
c.g.2
m3
WT(YT) = W1(Y1) + W2(Y2) + W3(Y3)
or
c.g.3
X3
X2
X1
mT(YT) = m1(Y1) + m2(Y2) + m3(Y3)