PROBLEM 2.56 The cabin of an aerial tramway is suspended from a set of wheels that can roll freely on the support cable ACB and is being pulled at a constant speed by cable DE. Knowing that 42α = ° and β = 32°, that the tension in cable DE is 20 kN, and assuming the tension in cable DF to be negligible, determine (a) the combined weight of the cabin, its support system, and its passengers, (b) the tension in the support cable ACB.

SOLUTION

Free-Body Diagram

First, consider the sum of forces in the x-direction because there is only one unknown force:

( ) ( )0: cos32 cos 42 20 kN cos 42 0x ACBF TΣ = ° − ° − ° =

or

0.1049 14.863 kNACBT =

(b) 141.7 kNACBT =

Now

( ) ( )0: sin 42 sin 32 20 kN sin 42 0y ACBF T WΣ = ° − ° + ° − =

or

( )( ) ( )( )141.7 kN 0.1392 20 kN 0.6691 0W+ − =

(a) 33.1 kNW =

56

PROBLEM 2.57 A block of weight W is suspended from a 500-mm long cord and two springs of which the unstretched lengths are 450 mm. Knowing that the constants of the springs are kAB = 1500 N/m and kAD = 500 N/m, determine (a) the tension in the cord, (b) the weight of the block.

SOLUTION

Free-Body Diagram At A

First note from geometry:

The sides of the triangle with hypotenuse AD are in the ratio 8:15:17.

The sides of the triangle with hypotenuse AB are in the ratio 3:4:5.

The sides of the triangle with hypotenuse AC are in the ratio 7:24:25.

Then:

( )AB AB AB oF k L L= −

and

( ) ( )2 20.44 m 0.33 m 0.55 mABL = + =

So:

( )1500 N/m 0.55 m 0.45 mABF = −

150 N=

Similarly,

( )AD AD AD oF k L L= −

Then:

( ) ( )2 20.66 m 0.32 m 0.68 mADL = + =

( )1500 N/m 0.68 m 0.45 mADF = −

115 N=

(a)

( ) ( )4 7 150: 150 N 115 N 05 25 17x ACF TΣ = − + − =

or

66.18 NACT = 66.2 NACT =

57

PROBLEM 2.57 CONTINUED

(b) and

( ) ( ) ( )3 24 80: 150 N 66.18 N 115 N 05 25 17yF WΣ = + + − =

or 208 N=W

58

PROBLEM 2.58 A load of weight 400 N is suspended from a spring and two cords which are attached to blocks of weights 3W and W as shown. Knowing that the constant of the spring is 800 N/m, determine (a) the value of W, (b) the unstretched length of the spring.

SOLUTION Free-Body Diagram At A

First note from geometry:

The sides of the triangle with hypotenuse AD are in the ratio 12:35:37.

The sides of the triangle with hypotenuse AC are in the ratio 3:4:5.

The sides of the triangle with hypotenuse AB are also in the ratio 12:35:37.

Then:

( ) ( )4 35 120: 3 05 37 37x sF W W FΣ = − + + =

or

4.4833sF W=

and

( ) ( )3 12 350: 3 400 N 05 37 37y sF W W FΣ = + + − =

Then:

( ) ( ) ( )3 12 353 4.4833 400 N 05 37 37

W W W+ + − =

or

62.841 NW =

and

281.74 NsF =

or

(a) 62.8 NW =

59

PROBLEM 2.58 CONTINUED

(b) Have spring force

( )s AB oF k L L= −

Where

( )AB AB AB oF k L L= −

and

( ) ( )2 20.360 m 1.050 m 1.110 mABL = + =

So:

( )0281.74 N 800 N/m 1.110 mL= −

or 0 758 mmL =

60