Question

the beam is uniform and weighs 60N. if 200N, find the tension in the tie rope x and y components of the force that the hinge exerts on the wall.

Answer 1

see arranged figure,

we have to find torque about the hinged point,

torque due to weight of beam + torque due to weight of object + torque due to tension = 0

or, $-60\times\frac{L}{2}cos30^{\circ}-200\times Lcos30^{\circ}+T\times(0.8Lcos60^{\circ})=0$

or, -60 × 0.5L × √3/2 - 200 × L × √3/2 + T × 0.8 × L × 1/2 = 0

or, -15√3L - 100√3L + 0.4TL = 0

or, 0.4T = 115√3 = 200

or, T = 500N

now, $\displaystyle\Sigma{F_x}$

= -T + H = 0 [ see figure ]

H = 500N

and $\displaystyle\Sigma{F_y}$

= -60 - 200 + V = 0

V = 260N

hence, horizontal component of force is 500N and vertically component is 260N

Answer 2

Answer:

Explanation:

Hey there,

The answer is following;

Data:

Weight: 60N

Tension: ?

Weight: 200N

Solution:

The data suggest that the torque of the object is to be found about the hinged point,

Torque(T) due to weight of beam + torque due to weight of object + torque due to tension = 0

Therefore,

-60 × 0.5L × √3/2 - 200 × L × √3/2 + T × 0.8 × L × 1/2 = 0

, -15√3L - 100√3L + 0.4TL = 0

, 0.4T = 115√3 = 200

T = 500N

now,

= -Torque + H = 0 [ see figure ]

x axis = 500N

and

= -60 - 200 + V = 0

y axis = 260N

hence, the hinge exerts 500N on the x-axis and 260N on the y axis of the wall.