An 8kg object is subjected to three forces.... F1=20i+30jN.... F2=22i-10jN...... F3=6i+4j.... a) Find the acceleration of the object........ b) If the object starts from rest from origin, what will be the location after 4sec.....c) What is the magnitude if resultant froce & its direction
Answers
Answer:
Resultant will be the sum of F1+F2.
Explanation:
F1=20i+30jN -1
F2=22i-10jN -2
F3=6i+4jN -3
1+2+3
F = 20i + 30j + 22i - 10j + 6i + 4j
F = 48i + 24j
F = √ (48)^² + (24)^²
F = √ 2304 + 576
F = √2880
F = 53.6656 Newton
a). F = mass * acceleration
53.6656 = 8 * acceleration
acceleration = 53.6656 / 8
acceleration = 6.7082 m/sec²
b). using 2nd law of equation
s = ut + 1/2 at²
s = 0*4 + 1/2(6.7082)(4)²
s = 0 + 53.6656
s = 53.6656 m
c). F1 = √(20)²+(30)²=√400+900=√1300
= 36.05
F2 = √(22)²+(-10)²=√484+100=√584
= 24.17
F3 = √(6)²+(4)² =√36 +16 =√52
= 7.21
tan Q = 36.05/24.17
= 1.49
= 1.49/7.21
= 0.21
= 1/√23
where Q = 12°