Homeworks
week 4
chapter 5  pg. 101 - 2,9,17,18,49,63,70,74
2
F1+F2 = 3i +4j + (-2)i + (-6)j = (3-2)i +(4-6)j = 1i -2j = m (axi + ayj) = 1 (axi + ayj) = axi + ayj
net force is 1i -2j
vectors are equal if x and y components are equal so:
1i -2j = axi + ayj      makes      ax= 1      ay= -2
size of acceleration vector =  and angle = tan-1( ay/ax)
9
The force vectors must be written in your coordinate system as in fig above.
Now you have to add all forces: (0i+5j)+(-11i+0j)+(0i-17j)+(3i+0j)+(12.12i+7j) = (0-11+0+3+12.12)i + (5+0-17+0+7)j = 4.12i-5j = net force (or total force) = according to Newton's Law = m (axi + ayj) = 4 (axi + ayj) =(4axi +4ayj)
vectors are equal if x and y components are equal so: 4.12 = 4ax     and    -5 = 4ay
so acceleration vector is 1.03i - 1.25j
size of acceleration vector =  and angle = tan-1( ay/ax)
17
    
from fig 2 you can find T:       +T-w = 5.5*0    (acceleration=0)   so T = 53.9N

from fig 3 you can find that        total mass = (4.8 + m + 5.5)
so Newton's Law will be:  +199 -  (4.8 + m + 5.5)*9.8 = (4.8 + m + 5.5)*0  (acceleration=0)
Solve for m.
Draw FBD for two lower pieces.  Write Newton's law knowing that acceleration is zero and solve for tension.
49

write Newton's Law for x direction and solve for acceleration.

write Newton's Law for x direction (you already know acceleration) and solve for M.
63

 

FBD shows the forces and their components in the coordinate system (x,y) .
Add all forces:
(0i+Nj) + (-980i -980j) + (Fcos30i -Fsin30j) = (-980+Fcos30)i + (N-980-Fsin30)j
if velocity is constant then acceleration is 0 so net force
(-980+Fcos30)i + (N-980-Fsin30)j =  0i +0j
This is equivalent to two equations one for x one for y axis. Solve one for F and the second for N
70

 
In standard coordinate system (y pointing up) 
first FBD:        +F - Mg = M( - a )    F positive, Mg negative, a negative
second FBD    +F - (M-m)g = (M -m) a   F positive, (M-m)g negative, a positive
Eliminating F:  -Mg + (M-m)g = M( - a ) - (M -m) a      so    - mg = -2Ma + ma
and:   2Ma = m (g+a)  and m = 2Ma/(g+a)