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2023 WAEC Further Mathematics Theory (a) A see-saw pivoted at the middle is kept in balance by weights of Richard,...

Further Mathematics
WAEC 2023

(a) A see-saw pivoted at the middle is kept in balance by weights of Richard, John and Philip such that only Richard whose mass is 60 kg sits on one side. If they sit at distances 2       m , 3 m , and 4 m respectively from the pivot and Philip is 15 kg, find the mass of John.

(bi) A body of mass 12 kg rests on a rough plane inclined at an angle of 30º to the horizontal. The coefficient of friction between the body and the plane is \(\frac{2}{3}\). A force of magnitude P Newton acts on the body along the inclined plane. Find the value of P, if the body is at the point of moving:

down the plane;

[Take \(g = 10 ms ^{-2}\)]

(bii) A body of mass 12 kg rests on a rough plane inclined at an angle of 30º to the horizontal. The coefficient of friction between the body and the plane is \(\frac{2}{3}\). A force of magnitude P Newton acts on the body along the inclined plane. Find the value of P, if the body is at the point of moving:

up the plane;

[Take \(g = 10 ms ^{-2}\)]

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Explanation

(a) Open photo

∑ clockwise moments = ∑ anti-clockwise moments

= 3 x mJ + 4 x 15 = 60 x 2

= 3mJ + 60 = 120

= 3mJ = 120 - 60

= 3mJ = 60

\(∴mJ=\frac{60}{3}=20kg\)

(bi)Open photo

\(∑f_y=0==>N-mgcosθ=0\)

\(=N-12(10)cos30^o=0\)

\(=N=120cos 30^o\)

\(=N=60√3N\)

At the point of moving down,

\(∑f_x=0==>μN-P-mgsinθ=0\)

\(=\frac{2}{3}(60√3)-P-12(10)sin30^o=0\)

\(=40√3-P-60=0\)

\(=P=40√3-60\)

\(∴P=9.28N\)

(bii)Open photo

\(∑f_y=0==>N-mgcosθ=0\)

\(=N-12(10)cos 30^o=0\)

\(=N=120 cos 30^o\)

\(=N=60√3N\)

At the point of moving up,

\(∑f_x=0==>P-μN-mgsinθ=0\)

\(=P-\frac{2}{3}(60√3)-12(10)sin30^o=0\)

\(=P-40√3-60=0\)

\(=P=40√3+60\)

\(∴P=129.28N\)


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