Resolving Forces : In-Class Home Exercise
2000N
m = 50kg
30ᵒ
g = 10m/s2
µ = 0.5
Solution
N = mg = 50 x 10 = 500N
First, draw the free-body diagram :
FAPP = 2000N
30ᵒ
PSF = µN = 0.5 x 500 = 250N
W = mg = 50 x 10 = 500N
First, need to resolve the applied force (FAPP = 2000N) into it’s x and y components. Don’t even think at this point – just
draw in the resolved force vectors and then apply sin and cos :
N = mg = 50 x 10 = 500N
FAPP = 2000N
FAPPY
PSF = 250N
30ᵒ
FAPPX
W = mg = 50 x 10 = 500N
Now, apply sin and cos :
sin 30
=
opposite =
hypotenuse
FAPPY =
FAPP
FAPPY
2000
 FAPPY =
FAPP x sin 30 = 2000 x 0.5 = 1000N
sin 30
=
adjacent =
hypotenuse
FAPPX =
FAPP
FAPPX
2000
 FAPPX =
FAPP x cos 30 = 2000 x 0.866 = 1732N
Now, construct force table :
Force
X Component
Y Component
FAPP
PSF
N
W
Total
FAPPX = 1732N
-250N
0N
0N
1732N – 250N = 1482N
FAPPY = 1000N
0N
500N
-500N
1000N + 500N + -500N = 1000N
So, now we know that there is an overall force of 1482N to the right and 1000N upwards. We can draw the resolved
total forces in the X and Y direction and then use Pythagorus to resolve these forces into a vector force with direction :
1000N
R
ᵒ
1482N
To determine the magnitude of R :
Pythagorus : c2 = a2 + b2
 R2 = 14822 + 10002
R=
14822 + 10002
To determine the angle that R makes with the horizontal :
tan  = opposite =
adjacent
1000 = 0.675
1482

 = tan -1 (0.675)
So, the box is moved by a force of 1787.83N at an angle of 34ᵒ.
Using F = ma, we can determine the acceleration of the box :
1787.83 = 50 x a 
a = 1787.83 = 35.77m/s2
50
 R = 1787.83N