B
8KN
Task 2a (P1*)
30°
5KN
A
A 4KN
C 1m
45°
0.5m
3KN
D
0.5m 1m
Determine the magnitude, direction and sense of the resultant and equilibrant forces acting on the
above component and the perpendicular distance of their line of action from point A.
Solution:
A. Fv= F×cos×ϑ A. Fh= F×sin×ϑ B. Fh= F×cos×ϑ B. Fv= F×sin×ϑ
= 4×cos×45 = 4×sin×45 = 8×cos×30 = 8×sin×30
= 2.8KN = 2.8KN = 6.9KN = 4KN
Fh=2.8KN 4KN Fv=4KN
8KN
A B
Fv=2.8KN Fh=6.9KN
Forces Fh Fv Turning moments Turning moments
3KN +3 0 0 3 3KN -4.2 0
4KN +2.8 +2.8 4.2 0 4KN 6.9 -2
5KN 0 5 0 7.5 5KN 0 0
8KN -6.9 +4 3.4 2 8KN 0 -1.5
∑= -1..8
,To find theta:
ϑ=cos-1( )
=cos-1( )
=84.6°
FR=11.74KN
Use of Pythagoras theorem to find resultant force (R). Fv=11.8
FR=
=139.24-1.21
=√
=11.74
ϑ
Fh=-1.1
To find the turning moments i multiplied the length from the pivot and the force
Forces Fh Fv
3KN -1.5 0
4KN 0 -4.2
5KN 0 0
8KN -2 -6.9
-3.5 -11.1
∑m-14.6Nm
x=
x=
=-1.2
The arrow going downwards is the
equilibrant force, which has the
same values as the Resultant force
, Task 2b (P1*)
A crate of mass of 50-100 Kg is held in equilibrium by a force T acting parallel to the plane
as indicated. Determine using the resolution method the magnitude of the force T and the
normal reaction R ignoring the effects of friction.
R
P
80KG
30°
Gravity=9.81N
w = mg
w = 80x9.81
w = 784.8N
(To find "R" we use cos and to find "P" we use sin)
R= w×cos×ϑ P= w×sin×ϑ
R=784.8×cos×30 P=784.8×sin×30
R=679.65N P=392.4N
The force "P" is 392.4N and the reaction force "R" is 679.65N.
8KN
Task 2a (P1*)
30°
5KN
A
A 4KN
C 1m
45°
0.5m
3KN
D
0.5m 1m
Determine the magnitude, direction and sense of the resultant and equilibrant forces acting on the
above component and the perpendicular distance of their line of action from point A.
Solution:
A. Fv= F×cos×ϑ A. Fh= F×sin×ϑ B. Fh= F×cos×ϑ B. Fv= F×sin×ϑ
= 4×cos×45 = 4×sin×45 = 8×cos×30 = 8×sin×30
= 2.8KN = 2.8KN = 6.9KN = 4KN
Fh=2.8KN 4KN Fv=4KN
8KN
A B
Fv=2.8KN Fh=6.9KN
Forces Fh Fv Turning moments Turning moments
3KN +3 0 0 3 3KN -4.2 0
4KN +2.8 +2.8 4.2 0 4KN 6.9 -2
5KN 0 5 0 7.5 5KN 0 0
8KN -6.9 +4 3.4 2 8KN 0 -1.5
∑= -1..8
,To find theta:
ϑ=cos-1( )
=cos-1( )
=84.6°
FR=11.74KN
Use of Pythagoras theorem to find resultant force (R). Fv=11.8
FR=
=139.24-1.21
=√
=11.74
ϑ
Fh=-1.1
To find the turning moments i multiplied the length from the pivot and the force
Forces Fh Fv
3KN -1.5 0
4KN 0 -4.2
5KN 0 0
8KN -2 -6.9
-3.5 -11.1
∑m-14.6Nm
x=
x=
=-1.2
The arrow going downwards is the
equilibrant force, which has the
same values as the Resultant force
, Task 2b (P1*)
A crate of mass of 50-100 Kg is held in equilibrium by a force T acting parallel to the plane
as indicated. Determine using the resolution method the magnitude of the force T and the
normal reaction R ignoring the effects of friction.
R
P
80KG
30°
Gravity=9.81N
w = mg
w = 80x9.81
w = 784.8N
(To find "R" we use cos and to find "P" we use sin)
R= w×cos×ϑ P= w×sin×ϑ
R=784.8×cos×30 P=784.8×sin×30
R=679.65N P=392.4N
The force "P" is 392.4N and the reaction force "R" is 679.65N.
