Question

PLS HELP ME IN THIS PROBLEM! DETERMINE FIRST ALL THE INSTANT CENTERS BEFORE SOLVING THE PROBLEM. I NEED THE CORRECT ANSWERS AND SOLUTION. THANK YOU

PROBLEM : After drawing the mechanism into a convenient scale to fit into the paper, determine Velocity of point B (in/sec) and angular velocity of link c rad per sec using the 'direct method,' if the circular cam a has wa = 15 rad/sec clockwise. There is a rolling contact between the cam a and the roller link b. b B 3.50 1.00 с с d 04.00 3.00 135° 3.00 1.22 do d a Dimensions are in inches.

Answer 1

The I centers are showen in the image.

A 134 B D 114 3 4 73.950 4.414 1.22 113 4.242 D d А 2 3.5 B D A 1 | 12 1.22 53.705° 2.5 4.414 d 123 52.340 124 1.661 A P

1- fixed link.

2-Cam

3-roller.

4-rod of roller.

23
,
124
is at point P.

is the point at which the line through the 2 centers and the fixed axis interrsect.

23

124
is th point of intersection of lines
12

14
and
23

4
.

From triangle dAD ,

Trignometric relations are used to fing AD and angle A.

In trianglee BAD,

cosine rule is used.

BD2 = ABP + ADP – 2 * AD * AB * cos A.

3.52 = 2.52 + 4.414? – 2* 2.5 * 4.414 * cos A.

A=.

52.34%

From triangle Pad.

tan 52.34 = Pd/1.22

Pd=1.661.

From I center method.

12 21 وان 14 24 الا

=15rad/s.

U2

1.661/5.903 = /15

امی

=4.2206rad/s.(angular velocity of link c).

امی

VB/D
=$V_{B}$ =
امی
*BD=4.2206*3.5=14.77 in/s.

Kennedys theorem circle Total no of links =4 cam,fixed link,roller,rod of roller. 112 Here,only 124 is needed.so it is the point of intersection of 112 14 and 123134. 114 112114 4.242 in 112124 1.661in 124 113 is the intersection of 114134 and 112123. 114124 5.903in 113 123 113114 3.5in 134