The jib crane shown in Fig. 4–45 a is subjected to three coplanar forces.
Replace this loading by an equivalent resultant force and specify where the resultant’s line of action intersects the column AB and boom BC.

Question

Accepted Answer

Force Summation. Resolving the 250-lb force into x and y components and summing the force components yields

[latex]\underrightarrow{+} (F_{R})_{x} = ∑F_{x}; (F_{R})_{x} = - 250 lb (\frac{3}{5} ) - 175 lb = - 325 lb = 325 lb ←[/latex]

[latex]+ \uparrow (F_{R})_{y} = ∑F_{y}; (F_{R})_{y} = - 250 lb (\frac{4}{5} ) - 60 lb = - 260 lb = 260 lb \downarrow [/latex]

As shown by the vector addition in Fig. 4–45 b ,

[latex]F_{R} = \sqrt{(325 lb)^{2} + (260 lb)^{2}} = 416 lb [/latex]

[latex] heta = an ^{- 1} \left(\frac{260 lb}{325 lb} ight) = 38.7°[/latex]

Moment Summation. Moments will be summed about point A .
Assuming the line of action of [latex]\mathbf{F}_{R}[/latex] intersects AB at a distance y from A , Fig. 4–45 b , we have

[latex]⤿+ (M_{R})_{A} = ∑M_{A}; 325 lb (y) + 260 lb (0) [/latex]
[latex]= 175 lb (5 ft) - 60 lb (3 ft) + 250 lb (\frac{3}{5} ) (11 ft) - 250 lb (\frac{4}{5} ) (8 ft) [/latex]
[latex]y = 2.29 ft [/latex]

By the principle of transmissibility, [latex]\mathbf{F}_{R}[/latex] can be placed at a distance x where it intersects BC , Fig. 4–45 b . In this case we have

[latex]⤿+ (M_{R})_{A} = ∑M_{A}; 325 lb (11 ft) - 260 lb (x) [/latex]
[latex]= 175 lb (5 ft) - 60 lb (3 ft) + 250 lb (\frac{3}{5} ) (11 ft) - 250 lb (\frac{4}{5} ) (8 ft) [/latex]
[latex]x = 10.9 ft [/latex]

Question 4.18: The jib crane shown in Fig. 4–45 a is subjected to three cop...

Related Answered Questions

Determine the moment MAB produced by the force F in Fig. 4–23a, which tends to rotate the rod about the AB axis. ...

Verified Answer:

Determine the magnitude of the moment of force F about segment OA of the pipe assembly in Fig. 4–24a. ...

Verified Answer:

Replace the two couples acting on the pipe column in Fig. 4–33a by a resultant couple moment. ...

Verified Answer:

Determine the couple moment acting on the pipe shown in Fig. 4–32a. Segment AB is directed 30° below the x–y plane. ...

Verified Answer:

The granular material exerts the distributed loading on the beam as shown in Fig. 4–51a. Determine the magnitude and location of the equivalent resultant of this load. ...

Verified Answer:

A distributed loading of p = (800x) Pa acts over the top surface of the beam shown in Fig. 4–50a. Determine the magnitude and location of the equivalent resultant force. ...

Verified Answer:

Determine the magnitude and location of the equivalent resultant force acting on the shaft in Fig. 4–49a. ...

Verified Answer:

Replace the force system in Fig. 4–47a by an equivalent resultant force and specify its point of application on the pedestal. ...

Verified Answer:

Replace the force and couple moment system acting on the beam in Fig. 4–44a by an equivalent resultant force, and find where its line of action intersects the beam, measured from point O. ...

Verified Answer:

The slab in Fig. 4–46a is subjected to four parallel forces. Determine the magnitude and direction of a resultant force equivalent to the given force system, and locate its point of application on the slab. ...

Verified Answer: