Question

the length L of the uniform bar is 3.60 m and its weight is 180 N. Also, let the block's weight ws 330 N and angle θ-30°. The wire can maximum tension of 500 N. With the block placed at this maximum x, what is the horizontal component of the force on the bar from the hinge at A? N (right) N (up) 12 2 ENG1/2

Answer 1

Forces acting on the bar are,

i)Weight of bar $W_1=180\,N$ acting at $L/2=3.60/2=1.80\,m$ from hinge

ii) Weight of the box $W=330\,N$ acting at from hinge

iii) components of force $T\cos\theta$ along the bar towards left and $T\sin\theta$ perpendicular to bar and upwards acting at a distance of $L=3.60\,m$ from the hinge.

iv) components of hinge forces $F_x$ along the bar, towards right and $F_y$ upwards acting at the hinge.

Net horizontal force on the bar is zero. $F_x=T\cos\theta$ ------(1)

Net vertical force on bar is zero, $F_y+T\sin\theta=W+W_1$ ----------(2)

Torque about the hinge,

$W*x+W_1*L/2=T\sin\theta*L$

$330*x+180*3.60/2=500\sin30\degree*3.60$

$x=1.745\,m$

Maximu possible distance before wire breaks is $x=1.745\,m$

b)

Horizontal component of force at hinge   $F_x=T\cos\theta=500*\cos30\degree=433\,N$

c)

Vertical component of the foce at hinge $F_y=W+W_1-T\sin\theta=330+180-250=260\,N$