1. Statics Worksheet 1 Answers
1. A worker that has a mass of 100 kg is standing on scaffolding whose
supports are 5.0 m apart. He is standing on a uniform beam of
mass 50.0 kg.
(a) Find the force on each support.
5.0 m
Wt2 F1 F2
Wt1
Wt1 = m1 g = ( 100 kg )( 9.8 m/s2 ) = 980 N
Wt2 = m2 g = ( 50 kg )( 9.8 m/s2 ) = 490 N

2. 5.0 m
490 N F1 F2
980 N
Forces up = Forces down
F1 + F2 =
(1) F1 + F2 = 1470

3. 5.0 m
490 N F1 F2
980 N
(1) F1 + F2 = 1470
Clockwise torques = Counter-clockwise torques
τCW = τCCW
Need a pivot point about which to apply the torques;
can choose any point we wish;
choose the pivot point to be at the left end of the beam

4. 5.0 m
490 N F1 F2
980 N
(1) F1 + F2 = 1470
Clockwise torques:
from the beam and the worker
Counter-clockwise torques:
from F2
Note: F1 does not produce a torque,
since its torque arm is zero

5. 5.0 m
490 N F1 F2
980 N
(1) F1 + F2 = 1470
τCW = τCCW
τ = r F
( 2.5 m )( 980 N ) + ( 2.5 m )( 490 N ) = ( 5 m ) F2
3675 m N = ( 5 m ) F2
F2 = 735 N

6. 5.0 m
490 N F1 F2
980 N
(1) F1 + F2 = 1470
(1) F1 + F2 = 1470
F2 = 735 N
F = 1470
F1 = 735 N

7. (b) If the man moves to a point 2
(b) If the man moves to a point 2.0 m from the left end, what is the force
on each support?
5.0 m
2.0 m
490 N F1 F2
980 N
Forces up = Forces down
F1 + F2 =
(1) F1 + F2 = 1470

8. 5.0 m
2.0 m
490 N F1 F2
980 N
(1) F1 + F2 = 1470
τCW = τCCW
τ = r F
( 2 m )( 980 N ) + ( 2.5 m )( 490 N ) = ( 5 m ) F2
3185 m N = ( 5 m ) F2
F2 = 637 N

9. 5.0 m
2.0 m
490 N F1 F2
980 N
(1) F1 + F2 = 1470
(1) F1 + F2 = 1470
F2 = 637 N
F = 1470
F1 = 833 N

10. what is the force on each support?
(c) If a load of bricks of mass 400 kg is placed 2.0 m from the right end,
what is the force on each support?
Wt. = ( 40 kg )( 9.8 m/s2 ) = 392 N
5.0 m
2.0 m
2.0 m F1
392 N
490 N F2
980 N
Forces up = Forces down
F1 + F2 =
(1) F1 + F2 = 1862

11. 5.0 m
2.0 m
2.0 m F1
392 N
490 N F2
980 N
(1) F1 + F2 = 1862
τCW = τCCW
τ = r F
(2 m)(980 N) + (2.5 m)(490 N) + (3.0 m)(392 N) = (5 m) F2
4361 m N = (5 m) F2
F2 = 872 N

12. 5.0 m
2.0 m
2.0 m F1
392 N
490 N F2
980 N
(1) F1 + F2 = 1862
(1) F1 + F2 = 1862
F2 = 872 N
F = 1862
F1 = 990 N

13. and again 3.0 m from that end. At the end hangs a sign that
2. A cantilever of length 8.0 m and mass 1000 kg is supported at one end
and again 3.0 m from that end. At the end hangs a sign that
weighs 300 N. Find the force on each support.
Wt. = ( 1000 kg )( 9.8 m/s2 ) = N
8.0 m
3.0 m
EAT F2 F1
9800 N
300 N
Forces up = Forces down
F1 + F2 =
(1) F1 + F2 =

14. 8.0 m
3.0 m
EAT F2 F1
9800 N
300 N
Clockwise torques = Counter-clockwise torques
Clockwise torques:
from the beam and the sign
Counter-clockwise torques:
from F2

15. 8.0 m
3.0 m
EAT F2 F1
9800 N
300 N
τCW = τCCW
τ = r F
( 4 m )( 9800 N ) + ( 8.0 m )( 300 N ) = ( 3 m ) F2
m N = ( 3 m ) F2
F2 = N

16. 8.0 m
3.0 m
EAT F2 F1
9800 N
300 N
(1) F1 + F2 =
F2 = N
F =
F1 = N
F1 must pull down on the beam in order for
equilibrium to be established

17. 3. Find the forces in the diagram below. The system is in equilibrium.
2 m m m m F1
40 N F2
50 N
75 N
85 N
Forces up = Forces down
F1 + F2 =
(1) F1 + F2 = 250

18. (1 m)(50 N) + (3 m)(75 N) + (4 m )(40 N) + (5 m)(85 N) = (6 m) F2
2 m m m m F1
40 N F2
50 N
75 N
85 N
τCW = τCCW
τ = r F
(1 m)(50 N) + (3 m)(75 N) + (4 m )(40 N) + (5 m)(85 N) = (6 m) F2
860 m N = ( 6 m ) F2
F2 = 143 N

19. 1 m
2 m m m m F1
40 N F2
50 N
75 N
85 N
(1) F1 + F2 = 250
F2 = 143 N
F = 250
F1 = 107 N