Analyzing a Truck and Trailer as a Frame The truck in Fig. 6.39 is parked on a 10° slope. Its brakes prevent the wheels at B from turning, but the wheels at C and the wheels of the trailer at A can turn freely. The trailer hitch at D behaves like a pin support. Determine the forces exerted on the

Question

Analyzing a Truck and Trailer as a Frame

The truck in Fig. 6.39 is parked on a 10° slope. Its brakes prevent the wheels at B from turning, but the wheels at C and the wheels of the trailer at A can turn freely. The trailer hitch at D behaves like a pin support. Determine the forces exerted on the truck at B, C. and D.

Strategy

We can treat this example as a structure whose "members" are the truck and trailer. We must isolate the truck and trailer and draw their individual freebody diagrams to determine the forces acting on the truck.

Accepted Answer

Determine the Reactions at the Supports The reactions in this example are the forces exerted on the truck and trailer by the road. We draw the freebody diagram of the connected truck and trailer in Fig. a. Because the tires at B are locked, the road can exert both a normal force and a friction force, but only normal forces are exerted at A and C. The equilibrium equations are

[latex] \sum{F_{x}} = B_{x} - 80 \sin 10º - 14 \sin 10º = 0 , [/latex]

[latex] \sum{F_{y}} = A + B_{y} + C - 8 \cos 10º - 14 \cos 10º = 0 , [/latex]

[latex] \sum{M_{(point A)}} = 14 B_{y} + 25C + (6)(8 \sin 10º) - (4)(8 \cos 10º) + (3)(14 \sin 10º) - (22)(14 \cos 10º) = 0 .[/latex]

From the first equation we obtain the reaction [latex]B_{x} = 3.82 kip[/latex], but we can't solve the other two equations for the three reactions A, [latex] B_{y}[/latex], and C .

Analyze the Members We draw the free-body diagrams of the trailer and truck in Figs, b and c, showing the forces [latex]D_{x} and D_{y}[/latex] exerted at the hitch. Only three unknown forces appear on the free-body diagram of the trailer. From the equilibrium equations for the trailer,

[latex] \sum{F_{x}} = D_{x} - 8 \sin 10º = 0 , [/latex]

[latex] \sum{F_{y}} = A + D_{y} - 8 \cos 10º = 0 , [/latex]

[latex] \sum{M_{(point D)}} =(0.5)(8 \sin 10º ) + (12)(8 \cos 10º) - 16A = 0[/latex].

we obtain A = 5.95 kip, [latex]D_{x} = 1.39 kip, and D_{y} = 1.93 kip.[/latex] (Notice that by summing moments about D, we obtained an equation containing only one unknown force.)

The equilibrium equations for the truck are

[latex] \sum{F_{x}} = B_{x} - D_{x} - 14 \sin 10º = 0[/latex] ,

[latex] \sum{F_{y}} = B_{y} + C - D_{y} - 14 \cos 10º = 0[/latex] ,

[latex] \sum{M_{(point B)}} = 11C + 5.5 D_{x} - 2D_{y} + (3)(14 \sin 10º) - (8)(14 \cos 10º) = 0 . [/latex]

Using the known values of [latex] D_{x} , and D_{y}[/latex], we can solve these equations, obtaining [latex] B_{x} = 3.82 kip, and B_{y} = 6.69 kip , and C = 9.02 kip .[/latex]

Discussion

We were unable to solve two of the equilibrium equations for the connected J truck and trailer. When that happens, you can use the equilibrium equations j for the entire structure to check your results:

[latex] \sum{F_{x}} = B_{x} - 8 \sin 10º - 14 \sin 10º = 3.82 - 8 \sin 10º - 14 \sin 10º = 0[/latex] ,

[latex] \sum{F_{y}} = A + B_{y} + C - 8 \cos 10º - 14 \cos 10º = 5.95 + 6.69 + 9.02 - 8 \cos 10º - 14 \cos 10º = 0[/latex] ,

[latex] \sum{M_{(point A)}} = 14B_{y} + 25C + (6)(8 \sin 10º) - (4)(8 \cos 10º) + (3)(14 \sin 10º) - (22)(14 \sin 10º) = (14)(6.69) + (25)(9.02) + (6)(8 \sin 10º) - (4)(8 \cos 10º ) + (3)(14 \sin 10º) - (22)(14 \cos 10º ) = 0[/latex].

Question 6.9: Analyzing a Truck and Trailer as a Frame The truck in Fig. 6...

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