 # Document

```Ledge Problems
• 2 Types (Both Solved with Torque)
– Board held by a cable
55.0
0
We can solve this problem by adding torques.
We have three forces acting on the beam.
1. The weight of the beam, acts at the beams center (its CG), FB.
2. The tension in the cable, T.
3. And the force exerted by the wall on the beam, R.
(The wall is pushing the beam up and out.)
Pivot point: where the beam meets the wall.
Therefore R exerts no torque as its lever arm is zero.
We only have two torques, they add up to zero.
Torque #1 is exerted by the tension in the cable
Torque #2 is caused by the weight of the beam.
Only the vertical component of the tension causes torque :
 cable  B  0
r
Tr sin   FB    0
2
T cos  RX  0
TO FIND T ON CABLE
 cable  B  0
Tr sin   FB rB  0
Tr sin   FB rB
TO FIND FORCE (R) OF WALL
Rx:
Ry:
T cos  RX  0
RY  FB  T sin 
TENSION ON CABLE
The 2.4 m. long weightless beam shown in the figure is supported on the right by a
cable that makes an angle of 50o with the horizontal beam. A 32 kg mass hangs
from the beam 1.5 m from the pivot point on the left. Find the T on cable.
Tr sin   FM rM  0
FM rM
T
r sin 
x
m = 32 kg
(32kg )(9.8)(1.5m)
T
2.4 sin 50
470N
 256 N
T
1.84
RESULTANT X FORCE (Rx)OF WALL
RX  T cos
RESULTANT Y FORCE (Ry)OF WALL
RESULTANT Y FORCE (R)OF WALL
Ry
Rx
R  ( Rx ) 2  ( R y ) 2
The 2.4 m. long weightless beam shown in the figure is supported on the right by a
cable that makes an angle of 50o with the horizontal beam. A 32 kg mass hangs
from the beam 1.5 m from the pivot point on the left. Find the R of wall.
T  256 N
b)
y direction:
T sin   R y  0
R y  (256 N ) sin 50.0o
x
m = 32 kg
R y  196.1N
x direction:
T cos   Rx  0
(256 N ) cos 50.0  Rx  0
o
R y  314 N  196.1N  118 N
R  ( Rx ) 2  ( R y ) 2
Rx  164.6 N
R  (164.6 N ) 2  (118 N ) 2
 140.0 N
A uniform beam is supported by a stout piece of line as shown. The beam weighs 175 N.
The cable makes an angle of 75.0. Find (a) the tension in the cable and
(b) the force exerted on the end of the beam by the wall.
Tr sin   FB rB  0
T
FB r
2 r sin 

175 N
2 sin 75.0o

75.0
4.00 m
90.6 N
(b) Next we can sum up the forces:
x direction:
T cos  RX  0
T
RX  T cos
y direction:
75.0
  90.6 N  cos 75.0
 23.4 N
FB
RY  FB  T sin 
 175 N   90.6 N  sin 75.0o
R  Ry 2  Rx 2

R
87.5 N 2   23.4 N 2
 87.5 N

90.6 N
A uniform beam of weight 254 N sticks out from a vertical wall. A lightweight cable
connects the end of the beam to the wall, making an angle of 65.0 between the
beam and the cable. (a) What is the tension in the cable? (b) What is the force
exerted on the beam by the wall?
a)
Tr sin   FB rB  0
254 N
T
 140.1N
o
2 sin 65.0
FBeam
T
2sin 
b)
y direction:
T sin   R y  FB  0
R y  (254 N )  (140 N ) cos 65.0o
R y  127.1N
x direction:
T cos   Rx  0
R  Ry  Rx
2
2

(140 N ) cos 65.0o  Rx  0
Rx  59.17 N
127.1 N    59.17 N   140.0 N
2
2
Tr sin   FB rB
Find Rx
T cos   Rx  0
Find Ry
R y  FB  T sin 
Fm rm  FB rB  Tr sin   0
R x  T cos 
Tr sin   Fm rm  FB rB   0
TO FIND T ON CABLE
Tr sin   Fm rm  FB rB   0
TO FIND F ON WALL
R y  FM  FB  T sin 
R x  T cos 
R  ( Rx ) 2  ( R y ) 2
The 2.4 m. long beam (30kg) shown in the figure is supported on the right by a cable that
makes an angle of 50o with the horizontal beam. A 32 kg mass hangs from the beam
1.5 m from the pivot point on the left. A) Determine the cable tension needed to
produce equilibrium. b) the force exerted on the end of the beam by the wall.
Fm rm  FB rB  Tr sin   0
470 N  352
Fm rm  FB rB
T
 448 N
T
o
sin 50.0 (2.4)
O
sin r
b)
x
Find Rx
m = 32 kg
R

T
cos

X
T cos   Rx  0
o
R

(
448
N
)
cos
50
.
0
 288 N
x
Find Ry
T sin   R y  FB  Fm  0
R y  FB  Fm  T sin 
a)
R y  (294 N )  (313N )  (448 N ) sin 50.0
R y  264 N
o
R  ( Rx ) 2  ( R y ) 2
R  ( 288) 2  ( 264) 2
A beam is supported as shown. The beam is uniform and weighs 300.0 N and 5.0 m long.
A 635 N person stands 1.50 m from the building. (a) What is the tension in the cable?
(b) the force exerted on the end of the beam by the wall.
(a) Sum of torques:
 beam  man  cable  0
Fm rm  FB rB  Tr sin   0
Fm rm  FB rB
T
r sin 
(635 N )(1.5m)  (300 N )(2.5m)
T
(5m) sin 55o
55.0
0
R
T
55.0
FB
T  416 N
Fm
(b) the force exerted on the end of the beam by the wall.
Find Rx
RX  T cos
o
Rx  (416 N ) cos 55.0  238 N
T cos   Rx  0
Find Ry
Ry  FB  Fm  T sin 
T sin   R y  FB  Fm  0
o
R y  (300 N )  (635 N )  (416 N ) sin 55.0
R y  624 N
R  R y  Rx
2
2

 624 N 
2
  238 N 
2

668 N
``` # Embry-Riddle Aeronautical University Jacobs MA 243 Final Examination (Sample) 