11–1.
The scissors jack supports a load P. Determine the axial P
force in the screw necessary for equilibrium when the jack
is in the position u. Each of the four links has a length 2 L and
is pin-connected at its center. Points B and D can move
horizontally.
C D
SOLUTION A B
x = 2L cos u, dx = - 2L sin u du u
y = 4L sin u, dy = 4L cos u du
dU = 0; - Pdy - Fdx = 0
- P(4L cos u du) - F(- 2L sin u du) = 0
F = 2P cot u Ans.
,.
,.
, 11–4.
The spring has an unstretched length of 0.3 m. Determine C
the angle u for equilibrium if the uniform links each have a
mass of 5 kg.
0.6 m
θ θ
SOLUTION
B
Free Body Diagram: The system has only one degree of freedom defined by the 0.1 m D
independent coordinate u. When u undergoes a positive displacement du, only the k = 400 N/m
A E
spring force Fsp and the weights of the links (49.05 N) do work.
Virtual Displacements: The position of points B, D and G are measured from the
fixed point A using position coordinates xB , xD and yG , respectively.
xB = 0.1 sin u dxB = 0.1 cos udu (1)
xD = 210.7 sin u2 - 0.1 sin u = 1.3 sin u dxD = 1.3 cos udu (2)
yG = 0.35 cos u dyG = - 0.35 sin udu (3)
Virtual–Work Equation: When points B, D and G undergo positive virtual
displacements dxB , dxD and dyG , the spring force Fsp that acts at point B does
positive work while the spring force Fsp that acts at point D and the weight of link
AC and CE (49.05 N) do negative work.
dU = 0; 21 -49.05dyG2 + Fsp1dxB - dxD2 = 0 (4)
Substituting Eqs. (1), (2) and (3) into (4) yields
134.335 sin u - 1.2Fsp cos u2 du = 0 (5)
However, from the spring formula, Fsp = kx = 4003210.6 sin u2 - 0.34
= 480 sin u - 120. Substituting this value into Eq. (5) yields
134.335 sin u - 576 sin u cos u + 144 cos u2 du = 0
Since du Z 0, then
34.335 sin u - 576 sin u cos u + 144 cos u = 0
u = 15.5° Ans.
and u = 85.4° Ans.
The scissors jack supports a load P. Determine the axial P
force in the screw necessary for equilibrium when the jack
is in the position u. Each of the four links has a length 2 L and
is pin-connected at its center. Points B and D can move
horizontally.
C D
SOLUTION A B
x = 2L cos u, dx = - 2L sin u du u
y = 4L sin u, dy = 4L cos u du
dU = 0; - Pdy - Fdx = 0
- P(4L cos u du) - F(- 2L sin u du) = 0
F = 2P cot u Ans.
,.
,.
, 11–4.
The spring has an unstretched length of 0.3 m. Determine C
the angle u for equilibrium if the uniform links each have a
mass of 5 kg.
0.6 m
θ θ
SOLUTION
B
Free Body Diagram: The system has only one degree of freedom defined by the 0.1 m D
independent coordinate u. When u undergoes a positive displacement du, only the k = 400 N/m
A E
spring force Fsp and the weights of the links (49.05 N) do work.
Virtual Displacements: The position of points B, D and G are measured from the
fixed point A using position coordinates xB , xD and yG , respectively.
xB = 0.1 sin u dxB = 0.1 cos udu (1)
xD = 210.7 sin u2 - 0.1 sin u = 1.3 sin u dxD = 1.3 cos udu (2)
yG = 0.35 cos u dyG = - 0.35 sin udu (3)
Virtual–Work Equation: When points B, D and G undergo positive virtual
displacements dxB , dxD and dyG , the spring force Fsp that acts at point B does
positive work while the spring force Fsp that acts at point D and the weight of link
AC and CE (49.05 N) do negative work.
dU = 0; 21 -49.05dyG2 + Fsp1dxB - dxD2 = 0 (4)
Substituting Eqs. (1), (2) and (3) into (4) yields
134.335 sin u - 1.2Fsp cos u2 du = 0 (5)
However, from the spring formula, Fsp = kx = 4003210.6 sin u2 - 0.34
= 480 sin u - 120. Substituting this value into Eq. (5) yields
134.335 sin u - 576 sin u cos u + 144 cos u2 du = 0
Since du Z 0, then
34.335 sin u - 576 sin u cos u + 144 cos u = 0
u = 15.5° Ans.
and u = 85.4° Ans.
