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Rigid Bodies and rotatational motion

  • Thread starter in10sivkid
  • Start date
yea...i'm having trouble today with my HW lol

a sign weighing 400 N is suspended at the end of a uniform rod 4.00 m weighing 500 N. what is the tension in the support cable if it makes an angle 40 with the rod?

now i'm curious i don't think Tclockwise = Tcounterclockwise comes into play

so i thought of this as a typical static equilibrium problem, but there is only 1 unknown that is the T in the cable that needs to be broken up into a vector

so exactly how does this get setup then

sum of Fx = Cos40F1 = 500 N + 500 N?
sum of Fy = Sin40F1 = 500 N + 400 N?

i know i am missing something
 
help please?
 

OlderDan

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Homework Helper
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in10sivkid said:
yea...i'm having trouble today with my HW lol

a sign weighing 400 N is suspended at the end of a uniform rod 4.00 m weighing 500 N. what is the tension in the support cable if it makes an angle 40 with the rod?

now i'm curious i don't think Tclockwise = Tcounterclockwise comes into play

so i thought of this as a typical static equilibrium problem, but there is only 1 unknown that is the T in the cable that needs to be broken up into a vector

so exactly how does this get setup then

sum of Fx = Cos40F1 = 500 N + 500 N?
sum of Fy = Sin40F1 = 500 N + 400 N?

i know i am missing something
Torque does come into play in this problem. There are forces acting where the rod is connected to the wall that are unkown in addition to the tension being unknown. Draw a free body diagram for the rod.
 
Last edited:
so curious would the summation of the forces in the y direction be

Sin40T + F1 (force rod is connected to the wall) = 400 N + 500 N
and but then the sum of the torque is getting me mixed up

if i choose the point where the rod is connected to the wall as the axis of rotation do I include Sin40T as well?


...also i'm having a hard time figuring out exactly how to setup the Torque if i do that...some help would be great
 

OlderDan

Science Advisor
Homework Helper
3,021
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in10sivkid said:
so curious would the summation of the forces in the y direction be

Sin40T + F1 (force rod is connected to the wall) = 400 N + 500 N
and but then the sum of the torque is getting me mixed up

if i choose the point where the rod is connected to the wall as the axis of rotation do I include Sin40T as well?


...also i'm having a hard time figuring out exactly how to setup the Torque if i do that...some help would be great
Forget what I said (and deleted) about not needing the torque. It would apply only to a massless rod. Your equation for the vertical forces is correct. Choosing the point of contact of the rod and wall for the axis of rotation leaves you with three forces to consider for the torque calculation. This is the only equation you need. You would need the equation you wrote if you wanted to find the force the wall applies to the rod.

Calculate two clockwise torques: (1) the 400N sign at distance 4m and (2) the 500N weight of the rod at distance 2m (the center of the rod). Calculate the counterclockwise torque: the vertical component of tension Tsin40 at 4m. Equate the clockwise torques to counterclockwise torque and solve for T.
 

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