Question

An aircraft tow bar is positioned by means of a single hydraulic cylinder connected by a 25-mmdiameter steel rod to two identical arm-and-wheel units def. the mass of the entire tow bar is 200 kg, and its center of gravity is located at g. for the position shown, determine the normal stress in the rod.

Answer 1

hi.....

I didn't understood the question so can you ask it one more time or you can write it in the comment section

Answer 2

Answer: R = 2654.5 N; $\sigma_{CD}  = -4.97MPa$

For Free body the entire tow bar calculation

IMG - 1

$W = (200Kg) \times (9.81m/ s ^2)$

= 1962.00 N

$\Sigma M _A = 0:850R - 1150 \times (1962.00N) = 0$

R = 2654.5 N

For free body both arm and wheel units calculation

IMG - 2

$tan\alpha = 100 /675$

α = 8.4270°

$\Sigma M _E = 0:(F_{CD} cos\alpha) (550)-R(500) = 0$

$F_{CD}$ = 500 /550 cos8.427° (2439.5N)

=2439.5 N (comp.)

$\sigma_{CD} = - F_{CD} /A_{CD} = - 2439.5N /(\pi(0.0125m )^2 )</p><p>[tex]= -4.9697 \times 10{^6} Pa$

$\sigma_{CD} = -4.97MPa$