Triangle of Forces (Past questions)

1.(a)  (i)      1.     mass = 360 / 9.8 36.7 (kg)                                                            B1

        (allow 2sf)

(ii)     2.     density = mass / volume                                                                C1

                   = 36.7 / 4.7 ´ 10–3

                   = 7.8 ´ 10–3                                                                       A1

          unit kg m–3                                                                                  B1

(ii)     right angled triangle with an additional correct angle marked               M1

set of correct force labels and correct arrows                                       A1

algebra shown or scale given                                                                C1

tension = 270 (N) or value in the range 255 to 285 (N)                        A1

  (b)   (i)      tension is a vector / has magnitude and direction                                  B1

direction involved in addition / the tensions or ropes act in
different directions                                                                               B1

(ii)     sum =270 sin37 + 360 sin53                                                                 B1

       =162.5 + 287.5                                                                              B1

(or one mark each for values of 162.5 and 287.5 seen) = 450 (N)       A0

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  2.     (a)     (pulley wheel) at rest / in equilibrium / acceleration is zero                            B1

(b)     (i)      500 N force down and general shape correct                                        B1

angles correct (one angle labelled correctly)                                         B1

T1 and T2 directions labelled correctly                                                  B1

(ii)     Formulae correct (resolving or sin rule) / scale diagram drawn correctly

with scale given                                                                                    B1

T1 = 674 (N) allow 650 to 700 for scale diagram                                 A1

T2 = 766 (N) allow 740 to 790 for scale diagram                                 A1

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  3.     (a)     weight  = 28 × 9.8 / mg                                                                                   C1
            = 270 (N)      (274.4)                                                                          A1
(using g = 10 then –1)

(b)     a completed triangle drawn with correct orientation                                       B1
at least two labels for triangle with correct directions given                           B1

          calculation:                                    scale diagram:
force P / weight = tan 35                scale given                                               C1
force P =192 (N)                            185 to 200 (N)                                         A1

  (c)   tension is greater                                                                                            B1
(reference to triangle) tension force would be greater (longer) as the holding
force P would be larger (longer) for greater angle / larger
value needed so vertical component still balances the weight                        B1

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  4.     (a)     (i)      1 Horizontal component = 24cos30                                                      C1

                                     = 21    (20.8) (N)                                              A1

2. vertical component = 24sin30

                                 = 12        (12.0) (N)                                             A1

(ii)     vertical force = 65 + 12                                                                        M1

                             = 77                                                                                  A0

(iii)    horizontal force = 20.8 (note ecf for 20.8 component)

resultant = [(77)2 + (20.8)2]1/2                                                               C1

                          = 80 (79. 8) (N)                                                          A1

(or by vector triangle need correct labels and arrows for C1 mark)

(iv)    80 (79.8)(N) / equal to (iii)  allow ecf                                                   B1

the resultant force needs to be zero or forces need

to balance above value to give no acceleration or constant velocity     B1

  (b)   (i)      P = F / A                                                                                               C1

    = 77 / 4.2 ´ 10–3

    = 18000 (18333) (Pa)                                                                       A1

(ii)     more / increases

downward / vertical component (of P) will be greater                           B1

(for larger angles)

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