statics worksheet solutions. 1. a simple beam bridge is 50.0 m long and has a mass of 20.0 metric...
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Statics Worksheet Solutions
1. A simple beam bridge is 50.0 m long and has a mass of 20.0 metric tons. (a) Find the center of mass of the bridge. (b) If the bridge is supported by a pillar at each end, find the force that each pillar exerts in supporting the bridge. (Hint: 1 metric ton = 1000 kg. Finding the torque about each pillar gives only one unknown for each equation.)
F1F2
mg
Center of mass at the center of the beam, x = 25.0 m from left. From symmetry of forces F1 and F2 about the center, get F1 = F2. Use torque and a pivot point at the left end to get F1:
mkgmmgmFs
m 0.2580.9200000.250.50 22
NFF 000,9812
2. If a 5.00 metric ton truck sits 20.0 m from the left end of the bridge, find the center of mass and the force of support from each pillar.
F1F2
mg bridge
mg truck
20.0 m
ton
mtonmton
mm
xmxmxcm 25
2520205
21
2211
mxcm 0.24
2. {continued} Put pivot at left end to get the force F2 from torque:
mgmmgmmF bridgetruck 0.250.200.502
NF 600,1172
Use net force to get to F1:
gmgmFFF bridgetrucknet 210
NF 400,1271
3. Repeat #2 if a 7.50 metric ton truck sits 10.0 m from the right end of the bridge in addition to the 5.00 metric ton truck 20.0 m from the left.
F1F2
mg bridge
mg truck
20.0 m
321
332211
mmm
xmxmxmxcm
mxcm 7.27
10.0 m
mg other truck
ton
mtonmtonmtonxcm 5.32
405.72520205
3. {continued} Put pivot at left end to get the force F2 from torque:
mgmmgmmgmmF truckbridgetruck 0.400.250.200.502
NF 400,1762
Use net force to get to F1:
gmgmgmFFF truckbridgetrucknet 210
NF 100,1421
4. A 20.0 kg sign is supported from the end of a massless horizontal 2.00 m rod. Find the tension in the rope and the force that the wall exerts on the beam. The rope is attached to the far end of the beam and makes an angle of 20.0 degrees to the horizontal. (You will have to use both F = 0 and = 0)
T = force of tension
= 20.0o
mg sign
Fy
Fx
Put pivot at left end to get the tension force T from torque:
mgmmT signo 00.20.20sin00.2
NT 573
4. {continued} Use net force to get to Fx and Fy:
cos0 TFF xxnet
NTFx 539cos
mgTFF yynet sin0
0sin TmgFy
5. If the 2.00 m rod in #4 has a mass of 8.50 kg and a uniform density, find the tension in the rope and the force that the wall exerts.
T = force of tension
= 20.0o
mg sign
Fy
Fx
Put pivot at left end to get the tension force T from torque:
mgmmgmmT signbeamo 00.200.10.20sin00.2
NT 695
mg beam
5. {continued} Use net force to get to Fx and Fy:
cos0 TFF xxnet
NTFx 653cos
gmgmTFF beamsignyynet sin0
NTgmgmF beamsigny 7.41sin
6. A steel beam of mass 1000 kg and length L is supported by posts at each end. A second beam, of equal density but half its length, lies in top of the first beam. If the left end of the top beam lines up with the left end of the left end of the bottom beam, find the force of support in each post.
F1 F2mg
mg21
2L
4L
6. {continued} Put pivot at left end to get the force F2 from torque:
42 21
2
Lmg
LmgLF
mgmgmg
F8
5
822
Use net force to get to F1:
mgmgFFFnet 21
210
mgmgmgFmgmgF 85
21
221
1
mgF8
71
7. A 3.00 m beam extends horizontally from a wall and has a mass of 45.0 kg. A wire extends form the wall above the beam at an angle of 45.0 degrees and connects 1.00 meter from the end of the beam. Find the tension in the rope and the force that the wall exerts on the end of the beam.
T = force of tension
mg
Fy
Fx
7. {continued} Put pivot at left end to get the force F2 from torque:
mmgmT o 50.10.45sin00.2
NT 467
cos0 TFF xxnet
NTFx 331cos
mgTFF yynet sin0
NTmgFy 110sin
Use net force to get to Fx and Fy:
8. (a) A 5.00 m long ladder leans against a wall at a point 4.00 m above the ground. The ladder is uniform and has a mass of 12.0 kg. Assuming that the wall is frictionless (the ground is not frictionless), determine the forces exerted on the ladder by the ground and wall. {126 N @ 70o and 44 N} (b) A 60.0 kg painter stands on the ladder 3.00 meters along the ladder from the bottom. Find the force exerted on the ladder by the wall and the floor. (c) If the ladder just begins to slip at its base at this point, what is the coefficient of static friction between the floor and the ladder? {Hint: Use the x- and y-components of the force on the ladder by the floor.}
m00.4
m00.3
n = normal force by the wall
Fy
Fx
mg ladder
m
m
00.5
00.4sin
o1.53
o1.53
o9.36
8. {continued} Put pivot at bottom left end to get the force n from torque:
oo mmgmn 9.36sin50.21.53sin00.5
o
os
m
m
mkgn
1.53sin00.5
9.36sin50.280.90.12 2
Nn 15.44
nFF xxnet 0 NnFx 15.44
mgFF yynet 0 NmgFy 6.117
Use net force to get to Fx and Fy:
m00.4
m00.3
n = normal force by the wall
Fy
Fx
mg laddero1.53
o1.53
o9.36
(b) A 60.0 kg painter stands on the ladder 3.00 meters along the ladder from the bottom. Find the force exerted on the ladder by the wall and the floor. (c) If the ladder just begins to slip at its base at this point, what is the coefficient of static friction between the floor and the ladder? {Hint: Use the x- and y-components of the force on the ladder by the floor.}
mg painter
o9.36
8. {continued} Put pivot at bottom left end to get the force n from torque:
oladder
o mgmmn 9.36sin50.21.53sin00.5
Nn 309
opainter mgm 9.36sin00.3
nFF xxnet 0 NnFx 309
gmgmFF painterladderyynet 0
NFy 6.705
Use net force to get to Fx and Fy:
8c. Fy is a normal force and fx is a static friction force. The relation of the two forces is given by the definition:
nF ss
ysx FF
438.06.705
309
N
N
F
F
y
xs
9. A 5.00 m long, 20 kg beam extends out from a wall at an angle of 30.0 degrees above the horizontal. A horizontal wire reaches from the wall and attaches 1.00 m from the upper end of the beam. Find the force exerted on the beam by the wall and the tension in the wire
o0.30 mg
To0.30
o0.60
Fy
Fx
9. {continued} Put pivot at bottom left end to get the force T from torque:
obeam
o mgmmT 0.60sin50.20.30sin00.5
NNn 1707.169
TFF xxnet 0 NTFx 170
gmFF beamyynet 0 NFy 196
Use net force to get to Fx and Fy:
10. Add a 10.0 kg mass is added to the upper end of the beam. Find the force on the beam by the wall and the tension in the wire.
o0.30
gmbeam
To0.30
o0.60
Fy
Fx
gmload
o0.60
10. {continued} Put pivot at bottom left end to get the force T from torque:
obeam
o mgmmT 0.60sin50.20.30sin00.5
NNn 3405.339
TFF xxnet 0 NTFx 340
gmgmFF loadbeamyynet 0
NFy 294
Use net force to get to Fx and Fy:
oload mgm 0.60sin00.5
11. 115 kg Mr. Mosig stands at the pool-end of a 25.0 kg, 3.40 m long diving board. If the board is held to the ground by two posts, one at the end of the board and one 70.0 cm in from the end, find the forces acting on the board by each of the supports.
gmbeam gmload1F 2F
m700.0 m00.1 m70.1
11. {continued} Put pivot at bottom left end to get the force T from torque:
mgmmgmmF loadbeam 40.370.1700.02
NF 60692
gmgmFFF loadbeamynet 210
NFy 4697
Use net force to get to F1:
