Question 1187414
The forces exerted by cable {{{T}}} and hinge {{{H}}} can be decomposed into their vertical components, {{{T[v]}}} and {{{H[v]}}} , and their horizontal components, {{{T[h]}}} and {{{H[h]}}} .
As the strut is uniform, we can work as if all of its weight is applied at its midpoint, at {{{"4.0 m"/2="2.0 m"}}} from the hinge.
The strut (horizontal), wall (vertical), and cable form a 3-4-5 right triangle, so the tension on the cable and its components (forming a similar triangle) are in the same {{{T[v]:T[h]T=3:4:5}}} ratio.
We have 4 significant figures for the weight and only 2 significant figures for the distances, so I would not know how many significant figures to use in the results. Use however many you think your teacher wants to see.
 
BALANCING TORQUES:
{{{T[v](4.0m)=(400.0N)(4.0m)+(600.0N)(2.0m)}}}
{{{T[v](4.0m)=2800Nm}}} --> {{{T[v]=700.0N}}}
{{{T=(5/3)T[v]=(5/3)(700.0N)=1166.7N}}} and
{{{T[h]=(4/3)T[v]=(4/3)(700.0N)=933.3N}}}
 
BALANCING HORIZONTAL (COMPONENT) FORCES:
{{{H[h]=T[h]=933.3N}}} (opposing forces with the same magnitude)
 
BALANCING VERTICAL FORCES/COMPONENTS:
Upward forces {{{H[v]}}} and {{{T[v]=700.0N}}} balance the downward weights {{{400.0N}}} and {{{600.0N}}} , so
{{{H[v]+700.0N=400.0N+600.0N}}} --> {{{H[v]=300.0N}}}
The magnitude of the force exerted by the hinge is
{{{H=sqrt((933.3N)^2+(300.0N)^2)=980.3N}}}
The angle it makes with the horizontal strut, {{{alpha}}} is such that
{{{tan(alpha)=300.0N/"933.3 N"=0.32144}}} --> {{{alpha=17.8^o}}}
 
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