4. Forces and Vectors
From Mechanics
i+Fcos(90−
)j=Fcos
i+Fsin
j
Express each of the forces given below in the form a
(a)
(b)
Solution
(a)
i+20sin40
j
(b)
i+80sin30
j
Note the negative sign here in the first term.
Express the force shown below as a vector in terms of
Solution
i−28sin30
j
Note the negative sign in the second term.
Express the force shown below as a vector in terms of
Solution
i−50sin44
j
Note that here both terms are negative.
Find the magnitude of the force (4
Solution
The magnitude, FN , of the force is given by,
42+82=
80=8
94 N (to 3sf)
The angle,
=tan−1
48
=63
4
Find the magnitude and direction of the resultant of the four forces shown in the diagram.
Solution
Force Vector Form
20 N
i+20sin50
j
18 N
25 N
i−25sin20
j
15 N
i+15sin30
j
20cos50
−25cos20
−15cos30
i+
20sin50
−18−25sin20
+15sin30
j=−23
627i−3
730j
The magnitude is given by:
23
6272+3
7302=23
9 N (to 3sf)
The angle
=3
73023
627
=9
0