Answer to Question #186369 in Mechanics | Relativity for Elle

Question

. A carpenter and his assistant are to carry a uniform wooden beam, which is 6 m long and weighs 100 kg, on their shoulders. Both men are of the same height so that the beam is carried horizontally. However, the carpenter is the stronger of the two and wishes to bear 50% more of the weight of the beam than his assistant. If one end of the beam is placed on the assistant’s shoulder, how far from the other end of the beam should the carpenter put his shoulder? [7]

Answer 1

Balancing forces in y direction,

$A+C=mg$

According to question

$C=\dfrac{50}{100}A+A$

$C=1.5A$

Substituting the value of C in above equation,

$2.5A=100\times9.8$

$A=392\space N$

$C=588\space N$

Taking moment about A,

$(A\times0)-(mg\times3)+(C\times(6-x))=0$

$6-x=\dfrac{3mg}{C}$

$x=6-\dfrac{3\times980}{588}=6-5$

$x=1\space m$