Sample Solutions
AP Physics B

Hecht:  Physics with Algebra/Trig


Chapter

Problems
6 pg 200-204
#31, 33, 37, 39, 60

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#31

The board must first be placed in translational equilibrium; that is,
(A)  SFx = 0 SFrightSFleft

this requirement is not applicable since there are no horizontal forces

(B)  SFy = 0 SFupSFdown

(beam is massless) S + R = (90.72)(9.8)
                        533.8 + R = 889

The board must then be placed in rotational equilibrium; that is,
 St = 0 Stcw Stccw

pivot at Rocko: (533.8)(1.829) = (90.72)(9.8)(x)

#33

For translational equilibrium:
(A)  SFx = 0 SFrightSFleft

this requirement is not applicable since there are no horizontal forces

(B)  SFy = 0 SFupSFdown

(bridge is massless) A + B = 8000

For rotational equilibrium:
 St = 0 Stcw Stccw

pivot at B: A(L) = (8000)(¼L)

#37

For translational equilibrium:
(A)  SFx = 0 SFrightSFleft

this requirement is not applicable since there are no horizontal forces

(B)  SFy = 0 SFupSFdown

(beam is massless) S + P = 2(9.8) + 1(9.8)

For rotational equilibrium:
 St = 0 Stcw Stccw

pivot at P: 1(9.8)(0.4) = 2(9.8)(0.1) + (S)(0.2)

#39 (see book for diagram)

For translational equilibrium:
(A)  SFx = 0 SFrightSFleft

this requirement is not applicable since there are no horizontal forces

(B)  SFy = 0 SFupSFdown

top board: B + D = 480
bottom board: A + C = B

For rotational equilibrium:
 St = 0 Stcw Stccw

top board - pivot at B: 480(9) = D(12)
bottom board - pivot at A: B(8) = C(12)

#60

For translational equilibrium:
(A)  SFx = 0 SFrightSFleft

this requirement is not applicable since there are no horizontal forces

(B)  SFy = 0 SFupSFdown

Fup = S = 200 + 320 + 200

For rotational equilibrium:
 St = 0 StcwStccw

pivot at S: 320(x) = 200(2.5+y) + 200(y)
note: uniform boards have their center of gravity in the middle
therefore  x + y = 2.5

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