(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. I...
Question 1 Report
(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 \(g = 10 ms ^{-2}\)]
Answer Details
(a) 
∑ 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)
\(∑f_y=0==>N-mgcosθ=0\)
\(=N-12(10)cos30^o=0\)
\(=N=120cos 30^o\)
\(=N=60√3N\)
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)
\(∑f_y=0==>N-mgcosθ=0\)
\(=N-12(10)cos 30^o=0\)
\(=N=120 cos 30^o\)
\(=N=60√3N\)
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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