- #1
Rob K
- 33
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Hi guys I was trying not to post this as it is difficult to understand without the diagram, but I will try and give you a visual. Basically there are 2 pulleys attached together on the one pivot, the larger one is driven by a belt that is driven by a motor. The inner pulley is used to hoist the weight.
Figure 3 shows a block A attached to a hoisting mechanism which consists of two
drums of radii r1 = 0:4 m, r2 = 0:8 m, ¯xed at O and which have a combined mass
of m1 = 100 kg and a radius of gyration about O of kO = 0:5 m. If a constant force
P = 1:5 kN is applied by the power unit, which consists of a pulley of radius r3
attached to a motor, determine the vertical acceleration of block A. The cord from
the motor to the drum is at an angle 40± to the horizontal. Block A has a mass
m2 = 300kg.
I = mk2
+CCW [itex]\Sigma[/itex]Mo = Io[itex]\alpha[/itex]
+[itex]\uparrow[/itex][itex]\Sigma[/itex]Fy = m(ag)x
+CCW a = [itex]\alpha[/itex]r
a = [itex]\alpha[/itex]r
I = 100kg * (0.5m)[itex]^{2}[/itex] = 25kg.m2
[itex]\Sigma[/itex]Mo
T(0.4m) = (25 kg.m2[itex]\alpha[/itex]
+[itex]\uparrow[/itex][itex]\Sigma[/itex]Fy = m(ag)x
-300(9.81)N + 1500sin40N + T = - 300a
a = [itex]\alpha[/itex]0.4
I worked out [itex]\alpha[/itex] to be 5.56, or 24. something. Basically after 20 odd hours, I have given up, there is something clearly very wrong and missing, and between me and 2 other people we don't know where to go from here.
The answers are as follows
[[itex]\alpha[/itex] = 0:312 rads/s2CW, T = 2980:5 N, a = 0:125 m/s2.]
Please help.
Kind Regards
Rob K
Homework Statement
Figure 3 shows a block A attached to a hoisting mechanism which consists of two
drums of radii r1 = 0:4 m, r2 = 0:8 m, ¯xed at O and which have a combined mass
of m1 = 100 kg and a radius of gyration about O of kO = 0:5 m. If a constant force
P = 1:5 kN is applied by the power unit, which consists of a pulley of radius r3
attached to a motor, determine the vertical acceleration of block A. The cord from
the motor to the drum is at an angle 40± to the horizontal. Block A has a mass
m2 = 300kg.
Homework Equations
I = mk2
+CCW [itex]\Sigma[/itex]Mo = Io[itex]\alpha[/itex]
+[itex]\uparrow[/itex][itex]\Sigma[/itex]Fy = m(ag)x
+CCW a = [itex]\alpha[/itex]r
a = [itex]\alpha[/itex]r
The Attempt at a Solution
I = 100kg * (0.5m)[itex]^{2}[/itex] = 25kg.m2
[itex]\Sigma[/itex]Mo
T(0.4m) = (25 kg.m2[itex]\alpha[/itex]
+[itex]\uparrow[/itex][itex]\Sigma[/itex]Fy = m(ag)x
-300(9.81)N + 1500sin40N + T = - 300a
a = [itex]\alpha[/itex]0.4
I worked out [itex]\alpha[/itex] to be 5.56, or 24. something. Basically after 20 odd hours, I have given up, there is something clearly very wrong and missing, and between me and 2 other people we don't know where to go from here.
The answers are as follows
[[itex]\alpha[/itex] = 0:312 rads/s2CW, T = 2980:5 N, a = 0:125 m/s2.]
Please help.
Kind Regards
Rob K