A uniform beam of mass M is supported in a horizontal position by a pin and cabl
ID: 1261000 • Letter: A
Question
A uniform beam of mass M is supported in a horizontal position by a pin and cable as shown in the Figure. Given. M = 27.0 kg ; m = 65.0 kg ; x = 2.90 m ; L = 5.60 m ; and ? = 25.0.
a) How many forces are acting on the beam (treat the interaction at the pivot with the wall as 2 forces)?
b) What is in N m, the torque due to the mass of the beam? (Use counterclockwise as positive)
c) What is in N the tension of the cable?
d) What is in N, H, the horizontal component of the force exerted by the wall onto the beam? Assume the positive direction points towards the right.
e) What is in N, V, the vertical component of the force exerted by the wall onto the beam? Assume the positive direction points upwards.
Use 10.0 N/kg for g.
A uniform beam of mass M is supported in a horizontal position by a pin and cable as shown in the Figure. Given. M = 27.0 kg ; m = 65.0 kg ; x = 2.90 m ; L = 5.60 m ; and ? = 25.0 . a) How many forces are acting on the beam (treat the interaction at the pivot with the wall as 2 forces)? b) What is in N m, the torque due to the mass of the beam? (Use counterclockwise as positive) c) What is in N the tension of the cable? d) What is in N, H, the horizontal component of the force exerted by the wall onto the beam? Assume the positive direction points towards the right. e) What is in N, V, the vertical component of the force exerted by the wall onto the beam? Assume the positive direction points upwards. Use 10.0 N/kg for g.Explanation / Answer
forces acting: weight (W) Tension (T) and compponents Tx and Ty
you can look at the moments produced around the pin.
M= (27*9.8*5.6*0.5) + (65*9.8*5.6) - Ty = 0 solve for Ty(tension in y direction)
Ty=4348.08 N
use right triangle trigonometry
sin(27)=Ty/T ---> T=Ty/sin(27)
T=9489.36 N tension of the cable
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