*25. ssm A 1220-N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960-N crate hangs from the far end of the beam. Using the data shown in the drawing, find (a) the magnitude of the tension in the wire and (b) the magnitudes of the horizontal and vertical components of the force that the wall exerts on the left end of the beam.

 

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Problem 25

Free Body Diagram

Take sum of torques about the point touching the wall which eliminates the Vertical Wall force (V) and the horizontal force (P).

 

Since the system is in static equilibrium the angular acceleration is zero.  For the torque due to tension, the component of tension perpendicular to the beam is opposite to the angle tension makes with the beam.  This angle is the sum of 50°+30°=80°

 

 

 

For the torque due to the weight of the beam and the weight of the crate I am going to use perpendicular distances.

 

 

 

 

 

 

 

 

 

 

To get the wall forces use the sum of the forces in the x and y directions.  Not for x and y of Tension use the 50° angle as that is the angle T is to the x axis.

 

Again because system is in static equilibrium.

 

 

 

 

 

 

 

 

 

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This page last updated on January 11, 2020