Question

A 50 kg uniform square sign, 2.00 m on a side, is hung from a 3.00 m rod of negligible mass. A cable is attached to the end of the rod and to a point on the wall 4.00 m above the point where the rod is fixed to the wall. What is the tension in the cable?

Tries 0/10

What is the horizontal component of the force exerted by the wall on the rod?

Tries 0/10

What is the vertical component of the force exerted by the wall on the rod?

Answer 1

(a) Tension in the cable which will be given as :

weight of the square box, w = m g

w = (50 kg) (9.8 m/s²) 490 N

using a trigonometric identity in above figure -

tan = AB / BC (4 m) / (3 m)

= tan^-1 (1.33)

= 53.1 degree

we know that, torque = r F sin                                                                     { eq.1 }

= (1) (w/2) sin 90⁰ + (3) (w/2) sin 90⁰ - (3 T) sin (180 - )

where, = 0

then, we have

0 = (w/2) + (3w / 2) - (3 T) sin (180 - )

T = 2 w / 3 sin (180 - )                                                                             { eq.2 }

inserting the value of 'w' & '' in eq.2,

T = 2 (490 N) / 3 sin (180 - 53.1)⁰

T = (980 N) / 3 ( 0.7996)

T = 408.5 N

(b) The horizontal component of the force exerted by the wall on the rod which is given as :

using an equation,    F_x = T cos                                                                { eq.3 }

inserting the values in eq.3,

F_x = (408.5 N) cos 53.1⁰

F_x = (408.5 N) (0.6004)

F_x = 245.2 N

(c) The vertical component of the force exerted by the wall on the rod which is given as :

using an equation,     F_y = w - T sin                                                                      { eq.4 }

inserting the values in eq.4,

F_y = (490 N) - (408.5 N) sin 53.1⁰

F_y = (490 N) - (408.5 N) (0.7996)

F_y = (490 N) - (326.6 N)

F_y = 163.4 N