Horizontal force on a horizontal bar

  • #1
Homework Statement
A uniform horizontal bar of mass m and
length L = 1.59 m is held by a frictionless
pin at a wall. The opposite end of the strut is
supported by a cord with tension T at an angle θ. A block of mass 2 m is hung from thebar at a distance of 3/4 L from the pin. If the mass of the bar is mass m = 1.52 kg, find the magnitude of the horizontal component of the force of the wall acting on the bar
if the string makes an angle of 39.7◦ with the
The acceleration of gravity is 9.8 m/s
Answer in units of N.
Relevant Equations
Tnet = sum Ti
Tnet = 0 = Tcord-Twall-Tmass
TLsinθ= -(3/2)mL-(1/2)mL
T= (1.52)/sin(39.7) = 2.38N


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  • #2
If I am not mistaken, the symbol ##T## stands for the tension in the rope. That is not the horizontal force of the wall acting on the bar. The torque balance equation cannot give you that. You also need the force balance equations. Also, you omitted the acceleration of gravity ##g## from the weights.
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Likes Kumail Haider and MatinSAR
  • #3
Tnet = 0 = Tcord-Twall-Tmass
=0, but that does not lead to:
TLsinθ= -(3/2)mL-(1/2)mL
Then you seem to have read that as +(3/2)mL-(1/2)mL to arrive at
Also, your overuse of "T" is confusing. You have used it, as given, for the tension in the rope, as a base for subscripts for other forces (F would have been clearer) and for torque (try τ).