Problem #1
Someone 45 N style at the end of the door is 84cm wide. What is the torque if the force given (a) is perpendicular to the door, and (b) at an angle of 600 to the front door?Answer:
The formula for torque is:
τ = r x F = rFsinθ
So for an angle of 600:
τ = (0.84 m)(45 N) sin (600) = 32.7 Nm = 33 Nm
If the force is applied at an angle of 900 to the radius, the sin factor θ becomes 1, then the torque value is:
τ = rF = (0.84 m)(45 N) = 37.8 Nm = 38 Nm
Problem #2
A force F is carried out on a horizontal homogeneous stem as shown below!
The correct statement for the moment on the rod with the point P because of this style is. . . .
A. F(sin θ)d
B. F(sin θ)d/L
C. Fd/L
D. F(cos θ)d
E. F cos θ)d/L
Answer: A
Torque is the product of thrust with distance (arm force or moment arm) measured from the shaft and perpendicular to the force line of work, then from the image above d sin θ is the moment arm in question because it is perpendicular to F, the moment of inertia is working on the stem is as big as,
τ = d sin θ (F)
Problem #3
F1, F2, F3 and F4 forces work on ABCD bars as shown.
If the stem mass is ignored, then the moment force value of point A is ...
A. 15 m. N
B. 17 m. N
C. 8 m.N
D. 63 m.N
E. 68 m.N
Answer: B
The moment of force with respect to point A (axis A) is
τA = F1 x 0 + (–F2 x AB) + F3 x AC + (–F4 x AD)
τA = 10 N x 0 + (–4 N x 2 m) + 5 N x 3 m + (–4 N x 6 m) = –17 mN
Problem #4
Calculate the total torque acting on the wheel shaft shown below. Assume that a 0.40 mN friction torque is opposed to movement.
A. +1,1 Nm
B. + 1.3 Nm
C. - 1.4 Nm
D. - 1.5 Nm
E. + 2.0 Nm
Answer: C
Because what we agree on is that a clockwise force has a negative torque and a counter-clockwise force has a positive torque, so from the picture above we get the torque caused by the three forces above to the shaft is
τ = –18 N x 0.24 m + (–35 N x 0.12 m) + 28 N x 0.24 m = –1.8 mN
Because there is torque due to friction opposite the movement with a large 0.4 mN, the total torque acting on the shaft is
τ total = –1.8 mN + 0.4 mN = –1.4 mN
Problem #5
A free two-wheeled system rotates against an axis without friction through the center with the wheel and perpendicular to the plane of the paper. Four forces are carried out in the tangential direction of the edge as shown in the figure besides the resultant moment of the system with respect to the axis is. . . .
A. zero
B. FR
C. 2FR
D. 6FR
E. 9FR
Answer: C
from the picture above we get the torque caused by the four forces above to the shaft is
τ = F x 2R + (–3F x 2R) + 2F x R + 2F x 2R = 2FR
Problem #6
An 8k N force works at O, the origin of the coordinate system. torque to point (-2, 1) is. . . .
A. –8(i - 2j)
B. –8(2i - j)
C. 8(i - 2j)
D. 8(4i + 2j)
E. 8(i + 2j)
Answer: E
The concept of cross multiplication vector:
i x j = k; j x i = –k; i x i = 0
j x k = i; k x j = –i; j x j = 0
k x i = j; i x k = –j; k x k = 0
note: thrust force is given by F = 8k = (0i + 0 j + 8k) and moment arm, r = (–2,1) = (–2i + j + 0k), and because torque is cross multiplication (vector multiplication) between r and F then,
τ = r x F
τ = (–2i + j + 0k) x (0i + 0 j + 8k)
= 0 - 16 (–j) + 0 + 8 (–i)
= 16j + 8i
τ = 8 (i + 2j)
Problem #7
If the given triangle plate is fixed from the point O and can rotate around this point, find the total torque applied by the given forces.
A. F(sin θ)d
B. F(sin θ)d/L
C. Fd/L
D. F(cos θ)d
E. F cos θ)d/L
Answer: A
τ = d sin θ (F)
Problem #3
F1, F2, F3 and F4 forces work on ABCD bars as shown.
A. 15 m. N
B. 17 m. N
C. 8 m.N
D. 63 m.N
E. 68 m.N
Answer: B
The moment of force with respect to point A (axis A) is
τA = F1 x 0 + (–F2 x AB) + F3 x AC + (–F4 x AD)
τA = 10 N x 0 + (–4 N x 2 m) + 5 N x 3 m + (–4 N x 6 m) = –17 mN
Problem #4
A. +1,1 Nm
B. + 1.3 Nm
C. - 1.4 Nm
D. - 1.5 Nm
E. + 2.0 Nm
Answer: C
Because what we agree on is that a clockwise force has a negative torque and a counter-clockwise force has a positive torque, so from the picture above we get the torque caused by the three forces above to the shaft is
τ = –18 N x 0.24 m + (–35 N x 0.12 m) + 28 N x 0.24 m = –1.8 mN
Because there is torque due to friction opposite the movement with a large 0.4 mN, the total torque acting on the shaft is
τ total = –1.8 mN + 0.4 mN = –1.4 mN
Problem #5
A. zero
B. FR
C. 2FR
D. 6FR
E. 9FR
Answer: C
from the picture above we get the torque caused by the four forces above to the shaft is
τ = F x 2R + (–3F x 2R) + 2F x R + 2F x 2R = 2FR
Problem #6
An 8k N force works at O, the origin of the coordinate system. torque to point (-2, 1) is. . . .
A. –8(i - 2j)
B. –8(2i - j)
C. 8(i - 2j)
D. 8(4i + 2j)
E. 8(i + 2j)
Answer: E
The concept of cross multiplication vector:
i x j = k; j x i = –k; i x i = 0
j x k = i; k x j = –i; j x j = 0
k x i = j; i x k = –j; k x k = 0
note: thrust force is given by F = 8k = (0i + 0 j + 8k) and moment arm, r = (–2,1) = (–2i + j + 0k), and because torque is cross multiplication (vector multiplication) between r and F then,
τ = r x F
τ = (–2i + j + 0k) x (0i + 0 j + 8k)
= 0 - 16 (–j) + 0 + 8 (–i)
= 16j + 8i
τ = 8 (i + 2j)
Problem #7
If the given triangle plate is fixed from the point O and can rotate around this point, find the total torque applied by the given forces.
Answer:
We find the torques of the forces one by one by one and finelly sum them up considering their directions.
τ1 = F1.d1 sin θ = 2F. 0 = 0
τ2 = F2.d2 sin θ = 5F.3d = 15Fd
τ3 = F3.d3 sin θ = –7F.6d = –42Fd
Total torque:
τtotal = τ1 + τ2 + τ3 = 0 + (15Fd) + (–42Fd) = –27Fd to the clockwise direction
Problem #8
If the plate is fixed from the point O, find the net torque of the given forces.
Answer:
τ1 = F1.L1 sin θ = –2F. 3L = –6FL
τ2 = F2.d2 sin θ = 3F. L = 3FL
τ3 = F3.d3 sin θ = 4F. 0 = 0
τ4 = F4.d4 sin θ = –8F. 2L = –16FL
Total torque:
τtotal = τ1 + τ2 + τ3 + τ4 = –6FL + 3FL+ 0 + (–16FL) = –19FL to the clockwise direction
We find the torques of the forces one by one by one and finelly sum them up considering their directions.
τ1 = F1.d1 sin θ = 2F. 0 = 0
τ2 = F2.d2 sin θ = 5F.3d = 15Fd
τ3 = F3.d3 sin θ = –7F.6d = –42Fd
Total torque:
τtotal = τ1 + τ2 + τ3 = 0 + (15Fd) + (–42Fd) = –27Fd to the clockwise direction
Problem #8
If the plate is fixed from the point O, find the net torque of the given forces.
τ1 = F1.L1 sin θ = –2F. 3L = –6FL
τ2 = F2.d2 sin θ = 3F. L = 3FL
τ3 = F3.d3 sin θ = 4F. 0 = 0
τ4 = F4.d4 sin θ = –8F. 2L = –16FL
Total torque:
τtotal = τ1 + τ2 + τ3 + τ4 = –6FL + 3FL+ 0 + (–16FL) = –19FL to the clockwise direction
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