In this problem we have a heavy beam suspended horizontally with a cable at one end and the cable
is not attached all the way to the very end it's attached to a spot here two and a half meters from
the left end and then the beam is hooked up to a hinge at the left end
and one of our questions is to find the tension in the cable but another one of our questions is
to find the force exerted by the hinge and before i get started doing any of the math
i wanted to figure out qualitatively which way does that force point so we use static equilibrium
reasoning for this i know that the sum of all the forces in the x direction must equal zero
if that wasn't true the rod would take off accelerating horizontally but it's
stationary right so i see the tension force has a leftward component to it and that means that the
force exerted by the hinge must have a rightward component to it in order to cancel that out so i'm
just going to go ahead and draw that component in real quick and i'll call that hinge force
x all right then for my next piece of reasoning i'm going to have to get the force of gravity
in here on the center of mass of my beam and i have options here for how to do this reasoning
but what i'm going to do is look at the rotation axis located at the connection point of the cable
and the torque exerted by the tension there is zero so all i have to think about is the hinge
force and the force of gravity so what i'm going to use here is the fact that the sum
of all of the torques must be zero about any rotation axis i want because this thing has
no rotational acceleration around any point so i chose my rotation axis right
here and i can see that gravity is exerting a counterclockwise torque with respect to that axis
which means the hinge must be exerting a clockwise torque with respect to that axis
and the x component has nothing to do with that that exerts no torque at all but what it means
is that my y component of the hinge force must point up and i called that f h y so then i can put
together those two components of the hinge force and show it now as a vector something like this
and although i didn't state it clearly in the question my goal here is to find the
magnitude and the direction of this hinge force in other words i want to express it in polar form
all right so what i'm going to do is take the actual hinge as the rotation
axis for the first part of my calculation and that's where i'm trying to find the tension
and i'm going to call that rotation axis number one and i'm going to look at the sum of the
torques around the rotation axis so the counter clockwise and clockwise torques must be equal
and with respect to that axis gravity is exerting a clockwise torque and that's
going to be m g times the length of the lever arm while gravity acts at the center of mass
so that's at the one and a half meter mark
and then my counterclockwise torque is exerted by the tension in the cable
and it's just the perpendicular component of that so it's going to be a t sine 25 multiplied
by the distance to the rotation axis which is two and a half meters so plugging in mg that's 120
times 9.8 times one and a half and crunching it all in my calculator i get 1 670 newtons just
keeping three sig figs there so what about the hinge force what i'm going to do is go after
this component by component so first off remember the x component of the hinge force must be exactly
canceling the x component of the tension pulling to the left and so that's a quick one i can write
the x component of the hinge force just the magnitude of it here is going to be t cosine 25
that's 1670 cosine 25 degrees and i've got one component of the hinge force that's 1513 newtons
just keeping an extra digit because i'm not done getting the polar form of this thing yet
what about the y component so the way i'm going to go after this is by using the fact that the sum of
all the forces in the y direction must be zero so i have two upward forces there's the upward
component exerted by the hinge and there's the upward component of the tension and then those
two things have to be equal in magnitude to the downward force exerted by gravity so i have
the y component of the hinge force plus a t sine 25 is equal to mg
and i'll go ahead and solve for f h y
that's mg minus t sine 25 and i'll put some numbers in there that's 120 times 9.8
minus 1670 sine 25 and i come out with a y component of my hinge force equal to 470
newtons and i still need to find the polar form of the hinge force so now i'm going to take these
two components and smash them together in the standard way all right there's my standard picture
of adding the vector components vectorially so head to tail to build a right triangle
and then i'll get the magnitude of the hinge force using the pythagorean theorem
so that's the square root of the sum of the squares 15 13 squared plus 470 squared
and when i smash these together i get 1584 newtons and to 3 sig figs i guess 1 580. it's slightly
different than the answer i got earlier today and it's just because i did a little more rounding
along the way then finally we'll get the angle and that's the angle whose tangent is 470 over 15 13.
so that's approximately 17.3 degrees so i'll leave this to you
but suppose you wanted to find some kind of independent check on your work here
what i would suggest is you create a third rotation axis right at the center of mass
and make sure the clockwise torque exerted by the hinge force
gives you exactly the same number as the counterclockwise torque exerted by
the cable and then you can be 100 certain you've got all the numbers right