- #1
JohnnyS
- 14
- 2
Homework Statement
I have been set this question and I am struggling with parts b and c. I think I am nearly there but can't quite get over the line. Please could someone give me a nudge in the right direction.
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1. (a) For the mechanism shown in FIGURE 1 determine for the angle
θ = 45°:
(i) the velocity of the piston relative to the fixed point O (VBO)
(ii) the angular velocity of AB about point A (i.e. ωAB)
(iii) the acceleration of point B relative to A (aBA).
(b) Determine the value of the angle θ (measured from vertical) when:
(i) the velocity of point B = 0
(ii) the angular velocity of link AB a maximum.
(c) What is the maximum angular velocity of link AB?
The Attempt at a Solution
1i) VBO = 1.1 ms-1
ii) ωAB = 5.55 rads-1
iii) 13.6cm and 40.8ms-2
The above I am happy with my answers which is why I have not included all my calculations and diagrams. Below is my attempt at the remaining questions.
1b)i) When the velocity of B is 0 ms-1 the piston must be either fully extended or fully retracted.
Offset length = cos45° x 50mm
= 35.36mm
Rod length at full extension = 200mm + 50mm
= 250mm
sinØ1 = 35.36
250
= sin-1(0.14144)
Ø1 = 8.13°
Ø = 90° – Ø1
= 90 – 8.13
Ø = 81.87°
When the rod is fully retracted the angle will be 270° – 8.13° = 261.87°
1b)ii)The angular velocity of link AB will be at a maximum when it is perpendicular to either the fully retracted or fully extended rod. Ø = 90° + 8.13° = 98.13°
1b)iii)
ωAB = VBA VBA = SinØ x VAO
lAB = Sin(98.13°) x 1.571
= 1.555 = 1.555ms-1
0.2
ωAB = 7.76 rad s-1 VAO = 1.571ms-1
Ø = 98.13°
lAB = 200mm = 0.2m
I am not looking for answers here just a nod in the right direction as I feel that I may have strayed off path a little here.
Thanks in advance for any help.
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