Question

2. The crossed four-bar linkage has a constant crank angular velocity ?2 = 10 rad/s ccw. The dimensions and the results of the kinematic analysis are given below for the position shown. (a) Draw the free-body diagrams of all the links of the mechanism, (b) Find the reaction forces at all the pins and the driving torque at link 2. The gravity centers of link 2 and link 4 are located at O2 and O4 respectively. O2A = 6 in, O2O4 = 18 in, AB = 18 in, O4B = 6 in, AC = 24 in, AG3 = 12 in, w3 = 4 lb, IG2 = IG4 = 0.063 lb·in·s^2, IG3 = 0.497 lb·in·s^2 ?3 = 1.43 rad/s, cw; ?4 = 11.43 rad/s, cw aG3 = (25.92 i + 24.58 j) ft/s^2 ?3 = ?4 = 84.8 rad/s^2 ccw Fc = 30 lb

          2. The crossed four-bar linkage has a constant crank angular velocity ?2 = 10 rad/s ccw. The dimensions and the results of the kinematic analysis are given below for the position shown. (a) Draw the free-body diagrams of all the links of the mechanism, (b) Find the reaction forces at all the pins and the driving torque at link 2. The gravity centers of link 2 and link 4 are located at O2 and O4 respectively.
O2A = 6 in, O2O4 = 18 in, AB = 18 in, O4B = 6 in,
AC = 24 in, AG3 = 12 in, w3 = 4 lb,
IG2 = IG4 = 0.063 lb·in·s^2, IG3 = 0.497 lb·in·s^2
?3 = 1.43 rad/s, cw; ?4 = 11.43 rad/s, cw
aG3 = (25.92 i + 24.58 j) ft/s^2
?3 = ?4 = 84.8 rad/s^2 ccw
Fc = 30 lb

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2. The crossed four-bar linkage has a constant crank angular velocity ?2 = 10 rad/s ccw. The dimensions and the results of the kinematic analysis are given below for the position shown. (a) Draw the free-body diagrams of all the links of the mechanism, (b) Find the reaction forces at all the pins and the driving torque at link 2. The gravity centers of link 2 and link 4 are located at O2 and O4 respectively.
O2A = 6 in, O2O4 = 18 in, AB = 18 in, O4B = 6 in,
AC = 24 in, AG3 = 12 in, w3 = 4 lb,
IG2 = IG4 = 0.063 lb·in·s^2, IG3 = 0.497 lb·in·s^2
?3 = 1.43 rad/s, cw; ?4 = 11.43 rad/s, cw
aG3 = (25.92 i + 24.58 j) ft/s^2
?3 = ?4 = 84.8 rad/s^2 ccw
Fc = 30 lb

Added by Michele S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

07/30/2024

Step 1

### Part (a) - Drawing Free-Body Diagrams (FBDs) #### Step 1: Draw FBD for Link 2 Show more…

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Transcript

00:01 In the question we have that the crossed four bar linkages are given to us.

00:06 Four bar linkages are given to us.

00:11 The diagram given looks something like this.

00:13 This is the y -axis.

00:15 This represents the x -axis.

00:17 Now this is the point where a is connected and then at this point it is c.

00:26 Here is point b.

00:30 We are given that this represents the angle of 321 .8 degree.

00:36 This is g3.

00:39 This is 3 to 60 degree.

00:44 O2.

00:46 This point is connected till here by 4.

00:51 This is 04.

00:53 This is 261 .8 degree.

00:56 Here is force fc acting.

00:59 This is the given diagram.

01:00 Now we need to draw the free body diagram.

01:03 The first part of the question, pre -body diagrams for the linkage mechanisms, for all links of the mechanism.

01:15 So to draw according to the links, we have the free body diagram number one.

01:20 That would look something like this.

01:22 This is a and this represents c.

01:25 There is a force acting fc while here would be another force that would be acting.

01:32 This is 38 .2 degree.

01:36 This is 3...

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