3. In the given 20.0 meter slab and inclined 10 with the horizontal, the ends are resting on inclines of 35° incline at B and 25° at A, a semi-trailer truck is to pass through it with the wheel loads as shown. How Far (x, meters) shall be the front wheels located before motion impends, the angle of friction is 18 ° for all areas of contact. If the weight of the slab is to be neglected 20.00 m 24KN 40 KN 32 KN 16 KN 3.6 m 1.2m 4.0 m 25° 35° 10°
3. In the given 20.0 meter slab and inclined 10 with the horizontal, the ends are resting on inclines of 35° incline at B and 25° at A, a semi-trailer truck is to pass through it with the wheel loads as shown. How Far (x, meters) shall be the front wheels located before motion impends, the angle of friction is 18 ° for all areas of contact. If the weight of the slab is to be neglected 20.00 m 24KN 40 KN 32 KN 16 KN 3.6 m 1.2m 4.0 m 25° 35° 10°
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter7: Dry Friction
Section: Chapter Questions
Problem 7.49P
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Can you please simplify the solution for this problem. Kindly clarify also if I am wrong I think our prof forgot to put the cos(10) for the distance of the restultant. (For the solution below)
Transcribed Image Text:3. In the given 20.0 meter slab and inclined 10 °with the horizontal, the ends are resting on inclines of 35° incline at B
and 25° at A, a semi-trailer truck is to pass through it with the wheel loads as shown. How Far (x, meters) shall be
the front wheels located before motion impends, the angle of friction is 18 ° for all areas of contact. If the weight of
the slab is to be neglected
20.00 m
40 KN
32 KN
24KN
16 KN
3.6 m
4.0 m
1.2m
25°
35°
10°
Transcribed Image Text:175KN
125 KN
T 140KN
BC
ViR
125(3m)- 90(4m) - FG(4) -0
FG- 3.953 KNC
CE
1ns (6) +125(9)-HOLG) + 155.85()
+ 90 (2) - BC (G) =0
BC 408.385 KN T
FG 3
155.885
125(9)+175 (12)+ 155. 885(12) + 90 (4)
- CE
JON T G:
은 (3,953)+90-은 GH-0
YB
CE 642,951 KNC
GH = 112.674 KNC
4.71 m
Fs
AD
24
1/2
A108
-
73°
47
Ns
70
120
112
RA
8A
NA
sin 120 Sin 70
430
112 (K+4.171)- 15.76 I Ca5 43° 20 00S 10° +
15.761 Sin 43 (20 sin 10) = 0
1/2 x + 467 K52-227.035 +37,331 =0
X: -2.477
RA 15.781 KN
16 24
A.0
40 32
123
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