A loading car, as given in figure, is at rest on a track forming an angle of 25° with the vertical.
The gross weight of the car and its load is 5500 lb, and it is applied at a point 30 in. from the
track, halfway between the two axles. The car is held by a cable attached 24 in. from the track.
Determine the tension in the cable and the reaction at each pair of wheels
Answers
Answer:
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Answer:
. For convenience, we choose:
• x-axis parallel to the track
• y-axis perpendicular to the track
2. The reaction at each wheel is perpendicular to the track.
3. The tension force T in the cable is parallel to the track.
4. The 5500 lb weight can be resolved into components:
Wx = +5500 cos 25^○ = +4980 lb \\Wy = −5500 sin 25^○ = −2320 lbWx=+5500cos25○=+4980lbWy=−5500sin25○=−2320lb
Step 2: Equilibrium Equations
.
Taking the moments about A to eliminate T and R1 :
Ç ∑MA = 0 :Ç∑MA=0:
−( 2320 lb )(25 in.) − ( 4980 lb)( 6 in.) + R2 ( 50 in.) = 0−(2320lb)(25in.)−(4980lb)(6in.)+R2(50in.)=0
R2 = +1758 lb , R2 = 1758 lb .R2=+1758lb,R2=1758lb.
Taking the moments about B to eliminate T and R2 :
+ Ç ∑MB = 0+Ç∑MB=0
(2320 lb )( 25 in.) − ( 4980 lb )( 6 in.) − R1 ( 50 in.) = 0(2320lb)(25in.)−(4980lb)(6in.)−R1(50in.)=0
R1 = +562 lb , R1 = 562 lbR1=+562lb,R1=562lb
The value of T is found by:
+ \ ∑Fx = 0 : \\+4980 lb −T = 0+ ∑Fx=0:+4980lb−T=0
T = +4980 lb , T = 4980 lb ÇT=+4980lb,T=4980lbÇ
Verifying:
∑Fy = 0 : \\+562 lb + 1758 lb − 2320 lb = 0∑Fy=0:+562lb+1758lb−2320lb=0