Question

Required information A weight W= 4 kN is supported by a cable that passes over frictionless pulleys at D and F. The cable is attached to a winch at G, and cable segment DG is vertical. Member ABC is built in at A, and members ABC, DCEF, and BE are attached using pins at points B, C, and E. Neglect the weights of individual members. All dimensions are in meters. 0.2 -1.6 0.9 o. LKJI 2.8— P TOEB 7 Determine the internal forces acting on cross sections Hand /, located immediately to the left of point Fand to the right of point E, respectively. The internal forces acting on cross section Hare as follows: NH=- D KN, VH= C kN, and MH= C kN.m The internal forces acting on cross section /are as follows: N=- D KN, V= D kN, and Mi=- D KN.m

Please solve this. Thank you.

Answer 1

given

weight of the block is W = 4 k N

the tension in the cable is T = W

T = 4 k N

applying the equilibrium condition to the above section

$\Sigma$ F_N = 0

N_H = - T

T = 4 k N

N_H = - 4 k N  

$\Sigma$ F_v = 0

V_H = T

T = 4 k N  

V_H = 4 k N   

the moment acting at the cross section H is

$\Sigma$ M_H = 0  

M_H = 0  

IN IN

$\Sigma$ F_N = 0

N_I = - T

N_I = - 4 k N

$\Sigma$ F_V = 0

V_I = T

V_I = 4 k N

$\Sigma$ M_I = 0

M_I = - 4 x 2.8

M_I = - 11.2 k N