Question

Chapter 12, Problem 021 Your answer is partially correct Try again. 38.70 and ? 50.0°, find (a) The system in the figure is in equilibrium. A concrete block of mass 286 kg hangs from the end of the uniform strut of mass 66.9 kg. For angles the tension Tin the cable and the (b) horizontal and (c) vertical components of the force on the strut from the hinge. Strut Hinge (a) Number (b) Number (c) Number3 Click if you would like to Show Work for this question: Units TN Units TN Units Open Show Work

Answer 1

a)
let M = 286 kg, m = 66.9 kg
let L is the length of the strut.

angle made by the cable with strut = 180 - (130 + 38.7) = 11.3 degrees

As the strut is in equilibrium net force and net torque acting on it must be zero.

T*L*sin(11.3) - M*g*L*sin(90-50) - m*g*(L/2)*sin(90-50) = 0

T*sin(11.3) - M*g*sin(40) - 0.5*m*g*sin(40) = 0

T = (M*g*sin(40) + 0.5*m*g*sin(40))/sin(11.3)

= (286*9.8*sin(40) + 0.5*66.9*9.8*sin(40))/sin(11.3)

= 1.03*10^4 N <<<<<<<<-------Answer

b) let Fh and Fv are the horizontal and vertical components of force exerted by the hinge.

Apply, Fnetx = 0

Fh - T*cos(38.7) = 0

Fh = T*cos(38.7)

= 1.03*10^4*cos(38.7)

= 8038 N <<<<<<<<-------Answer

c) Apply, Fnety = 0

Fv - M*g - m*g - T*sin(38.7) = 0

Fv = M*g + m*g + T*sin(38.7)

= 286*9.8 + 66.9*9.8 + 1.03*10^4*sin(38.7)

= 9898 N <<<<<<<<-------Answer