Horizontal force on a horizontal bar

AI Thread Summary
The discussion centers on the calculation of forces acting on a horizontal bar, specifically addressing the tension in the rope (T) and its relation to other forces. It highlights the importance of including the gravitational force in the equations and clarifies that torque balance alone is insufficient for determining horizontal forces. The misinterpretation of the equations is noted, particularly in the transition from torque to tension calculations. Additionally, the overuse of the symbol "T" creates confusion, suggesting that clearer notation would enhance understanding. Accurate representation of forces and careful notation are essential for solving the problem correctly.
marjine
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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
horizontal.
The acceleration of gravity is 9.8 m/s
Answer in units of N.
Relevant Equations
Tnet = sum Ti
T=rFsinθ
T=Ia
Tnet = 0 = Tcord-Twall-Tmass
TLsinθ-2m(3/4L)-m(1/2)L
TLsinθ= -(3/2)mL-(1/2)mL
T=m/sinθ
T= (1.52)/sin(39.7) = 2.38N
 

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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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marjine said:
Tnet = 0 = Tcord-Twall-Tmass
TLsinθ-2m(3/4L)-m(1/2)L
=0, but that does not lead to:
marjine said:
TLsinθ= -(3/2)mL-(1/2)mL
Then you seem to have read that as +(3/2)mL-(1/2)mL to arrive at
marjine said:
T=m/sinθ
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).
 
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