APM2EA1
2017
Problems on a Single Particle in Static Equilibrium:
1. The forces acting on the particle are F , F 1 , F 2 and F 3 .
The equilibrium condition is
F + F 1 + F 2 + F 3 = 0.
Applying the equilibrium condition to the x-components:
−F2 cos θ − F3 sin 30◦ + F1 cos 15◦ = 0
√
√
6+ 2
1
F1
⇒ F2 cos θ = − F3 +
2
2 √ √
!
2
− 21 F3 + 6+
F1
−1
2
⇒ θ = cos
F2
√
= cos−1
⇒ θ = 53.02039◦
1
√
− 21 (450) + 6+
2
750
2
(700)
!
Applying the equilibrium condition to the y-components and using the
numerical value of θ, then
F + F3 cos 30◦ + F1 sin 15◦ − F2 sin θ = 0
⇒ F = 750 sin θ − 700 sin 15◦ − 450 cos 30◦
⇒ F = 28.25 N.
2. The forces in the system are T 1 , T 2 and W . The following figure defines
T 1 and T 2 :
The equilibrium condition is
T 1 + T 2 + W = 0.
Applying the equilibrium condition to the x-components:
−T2 + T1 cos 30◦ = 0
√
3
⇒
T1 = T2
2
2
⇔ T1 = √ T2 .
3
Applying the equilibrium condition to the y-components:
T1 sin 30◦ − W = 0
1
⇒ W = T1
2
1
⇔ W = √ T2 .
3
With the constraint of T1 ≤ 2200 N, then W ≤ 1100 N and with the
constraint of T2 ≤ 2000 N, then W ≤ 1154 N. Notice that if T2 = 2000
N, then T1 > 2200 N and so the cable will break. If T1 = 2200 N, then
T2 < 2000 N and so the maximum weight that the engine can be is
Wmax = 1100 N.
3. The forces acting on the system are T 1 , T 2 and W . The following figure
defines T 1 and T 2 :
2
The equilibrium condition is
T 1 + T 2 + W = 0.
Applying the equilibrium condition to the x-components:
−T2 cos 30◦ + T1 cos 45◦ = 0
√
3
1
⇒
T2 = √ T1
2
2
√
6
⇒ T1 =
T2 .
2
Applying the equilibrium condition to the y-components:
T1 sin 45◦ + T2 sin 30◦ − W
1
1
⇒ √ T1 + T2
2
2
√
3
1
T2 + T2
⇒
2√
2
3+1
⇔
T2
2
=
=W
=W
=W
2W
⇒ T2 = √
3+1
300
=√
3+1
⇒ T2 = 109.81 N
√ 6
300
√
⇒ T1 =
2
3+1
⇒ T1 = 134.49 N.
4. The forces acting on the system are T 1 , T 2 , T 3 and W . The following
figure defines T 1 , T 2 and T 3 .
3
Each tension is given as follows.
T 1 = T1 n̂1
DB
DB
= −T1 ŷ.
= T1
T 2 = T2 n̂2
= T2
DA
DA
T2
(2x̂ − ŷ + 2ẑ) .
3
T 3 = T3 n̂3
=
= T3
=
DC
DC
T3
(−3x̂ + 2ŷ + 6ẑ) .
7
Finally, the weight is given by W = −W ẑ, where W = 150 N. The equilibrium condition is
T1 + T2 + T3 + W = 0
Applying the equilibrium condition to the x-components:
2
3
T2 − T3 = 0
3
7
9
⇒ T2 =
T3 .
14
Applying the equilibrium condition to the y-components:
2
1
−T1 − T2 + T3 = 0
3
7
2
1
⇒ T1 = T3 − T2 .
7
3
Applying the equilibrium condition to the z-components:
2
6
T2 + T3 − W = 0
3
7
2 9
6
⇒
T3 + T3 = W
3 14
7
7
⇒ T3 = W
9
350
=
N
3
= 116.67 N.
Thus,
9
T2 =
14
and
2
T1 =
7
350
3
350
3
−
4
= 75 N
1
(75) = 8.33 N.
3
—————–end————
5