1. A force P is applied to one end of a beam with a length of 2 m. What is the magnitude of the chasmhóiminte? The axis of rotation at point A.
Ar a dtugtar:
Fórsa (F) = 10 N
Length of AB (rAB) = 2 m
Force F is perpendicular to the beam.
The lever arm (l) =rAB peaca 90o = (2 m)(1) = 2 m
Teastaíonn: The torque about the axis of rotation
Réiteach:
An chasmhóimint:
τ = F l = (10 N)(2 m) = 20 N m
The plus sign because the beam rotates counterclockwise rotation.
2. The length of a beam AB is 2 m and the magnitude of force F is 10 N. What is the magnitude of the torque? The axis of rotation at point A.
Ar a dtugtar:
Fórsa (F) = 10 N
Length of AB (rAB) = 2 m
The lever arm (l) =rAB peaca 60o = (2 m)(0.5√ 3) = √ 3 m
Teastaíonn: The torque about the axis of rotation
Réiteach:
An chasmhóimint:
τ = F l = (10 N)(√ 3 m) = 10√ 3 N m
The plus sign because the force F causes the beam rotates counterclockwise rotation.
3. The length of a beam is 2 m. The magnitude of F1 is 10 N and the magnitude of F2 is 15 N. Determine the net torque about the center of the beam.
The axis of rotation at the center of the beam.
Ar a dtugtar:
Fórsa 1 (F1) = 10 N
An achar between F1 and the center of beam (r1) = 1 m
The lever arm 1 (l1) =r1 peaca 90o = (1 m)(1) = 1 m
Fórsa F1 is perpendicular to the beam.
Fórsa 2 (F2) = 15 N
The distance between F2 and the center of the beam (r2) = 1 m
Fórsa F2 is perpendicular to the beam.
The lever arm 2 (l2) =r1 peaca 90o = (1 m)(1) = 1 m
Ag Teastáil: The net torque about the axis of rotation
Réiteach:
The torque 1 :
τ1 = F.1 l1 = (10 N)(1 m) = 10 N m
The plus sign because the force of F1 causes the beam rotates counterclockwise rotation.
The torque 2 :
τ2 = F.2 l2 = (15 N)(1 m) = -15 N
The minus sign because the force F2 causes the beam to rothlaíonn deiseal.
The net torque :
Στ = τ1 – τ2 = 10 – 15 = – 5 N m
The minus sign because the beam to rothlaíonn deiseal.
4. The length beam AB is 2 m, The magnitude of F1 is 10 N and the magnitude of F2 is 10 N. Determine the net torque about the center of the beam.
The axis of rotation at the center of the beam.
Ar a dtugtar:
Fórsa 1 (F1) = 10 Nn
The distance between F1 and the center of the beam (r1) = 1 m
The lever arm 1 (l1) =r1 peaca 60o = (1 m)(0.5√ 3) =0.5√ 3 m
Fórsa 2 (F2) = 10 N
The distance between F2 and the center of the beam (r2) = 1 m
Fórsa F2 is perpendicular to the beam.
The lever arm 2 (l2) = r2 peaca 90o = (1 m)(1) = 1 m
Ag Teastáil: The net torque about the axis of rotation
Réiteach:
The torque 1 :
τ1 = F.1 l1 = (10 N)(0.5√ 3 m) = 5√3 = 8.7 N.m
The plus sign because the force F1 causes the beam to rothlaíonn deiseal.
The torque 2 :
τ2 = F.2 l2 = (10 N)(1 m) = -10 N m
The minus sign because the force F2 causes the beam to rothlaíonn deiseal.
The net torque :
Στ = τ1 – τ2 = 8.7 – 10 = – 1.3 N m
The minus sign because the net force causes the beam to rothlaíonn deiseal.
5. The length of a beam is 10 m, the magnitude of F1 is 10 N, the magnitude of F2 is 10 N and the magnitude of F3 is 15 N. The distance between point A and point C is 7.5 m. The Force F2 located at the center of the beam. Determine the net torque about the point C located at 2.5 m from the point B.
The axis of rotation located at point C
Ar a dtugtar:
Fórsa 1 (F1) = 10 N
The distance between F1 and point C (r1) = 2.5 m
Fórsa F1 is perpendicular to the beam.
The lever arm 1 (l1) =r1 peaca 90o = (2.5 m)(1) = 2.5 m
Fórsa 2 (F2) = 10 N
The distance between F2 and point C (r2) = 2.5 m
An fórsa F2 is perpendicular to the beam.
The lever arm 2 (l2) = r2 peaca 90o = (2.5 m)(1) = 2.5 m
Fórsa 3 (F3) = 15 N
The distance between F3 agus pointe C (r3) = 7.5 m
An fórsa F3 is perpendicular to the beam.
The lever arm 3 (l3) = r3 peaca 90o = (7.5 m)(1) = 7.5 m
Ag Teastáil: The net torque about the axis of rotation
Réiteach:
The torque 1 :
τ1 = F.1 l1 = (10 N)(2.5 m) = 25 N m
The plus sign because the force F1 causes the beam to rothlaíonn deiseal.
The torque 2 :
τ2 = F.2 l2 = (10 N)(2.5 m) = 25 N m
The plus sign because the force F2 causes the beam to rothlaíonn deiseal.
The torque 3 :
τ3 = F.3 l3 = (15 N)(7.5 m) = -112..5 N m
The minus sign because the force F3 causes the beam to rothlaíonn deiseal.
The net torque :
Στ = τ1 + τ2 – τ3 = 25 + 25 - 112.5 = – 62.5 N m
The minus sign because the net force causes the beam to rothlaíonn deiseal.
6. The length of a beam is 10 m, the magnitude of F1 is 10 N, the magnitude of F2 is 10 N and the magnitude of F3 is 10 N. Determine the net torque about point A, located 5 m from the poi
nt of application of force F1.
The axis of rotation at point A.
Ar a dtugtar:
Fórsa 1 (F1) = 10 N
The distance between F1 and point A (r1) = 5 m
The lever arm 1 (l1) =r1 peaca 60o = (5 m)(0.5√ 3) =2.5√ 3 m
Bhfeidhm 2 (F2) = 10 N
The distance between F2 and point A (r2) =0
An fórsa F2 is perpendicular to the beam.
The lever arm 2 (l2) = r2 peaca 90o = (0)(1) = 0
The force 3 (F3) = 10 N
The distance between F3 and point A (r3) = 10 m
The lever arm 3 (l3) = r3 peaca 30o = (10 m)(0.5) = 5 m
Ag Teastáil: the net torque about the axis of rotation
Réiteach:
The torque 1 :
τ1 = F.1 l1 = (10 N)(2.5√3 m) = 25√3 = 43.3 N m
The plus sign because the force F1 causes the beam to rothlaíonn deiseal.
The torque 2 :
τ2 = F.2 l2 = (10 N)(0) = 0
The torque 3 :
τ3 = F.3 l3 = (10 N)(5 m) = -50 N m
The minus sign because the force F3 causes the beam to rothlaíonn deiseal.
The net torque :
Στ = τ1 + τ2 – τ3 = 43.3 + 0 - 50 = – 6.7 N m
The minus sign because the net force causes the beam to rothlaíonn deiseal.
