Question

3. Consider again a two-link planar robot. Suppose that li = 5 m and l. = 3 m, 0 = 60°, and 82 = -75. a) Draw a diagram for the robot clearly showing the orientation of the links. vb) Obtain the vector for the location of the end-point of link 1 in rectangular coordinates. Call this vector 1. c) Obtain the vector in rectangular coordinates) representing the location of the end effector (end-point of link 2) relative to the end-point of link 1. Call this vector 12. - d) Obtain the vector for the location of the end-effector P. e) Write P in polar coordinates.

Answer 1

SOLUTION:

a)

759 BR

b)

OA= (1 = l1(cosi + sinouj)

li = 5(cos60° i + sin60°;)

4 . 33 + 2 . 3 = ا ج

c)

AB = 12 = 12(costzi + sintuj)

= =3(cos(-75°)i + sin(-75%)

2 . 808 - 0 . 776 = داچ

d)

first we write vector AB in the global frame:

AB = 12(cos(@1 + 02)i + sin(01 +02))

→ AB = 3(cos(60 – 75)i + sin(60 - 75);)

$\small OB = \vec{P} = \vec{l_1}+\vec{AB}$

$\small \Rightarrow \vec{P} = 2.5i+4.33j+2.898i-0.776j$

→ P=5.3981 +3.5547

e)

For polar co-ordinates, we find the length of the vector and the angle it makes.

$\small \left |\vec{P} \right |= \sqrt{5.398^2+3.554^2} = 6.463$

$\small \theta = tan^{-1}(\frac{3.554}{5.398}) = 33.361^0$

So the vector can be written in polar form as:

P=(,0) = (6.463, 33.361°