Asked by LEBZA
A minimum horizontal force of 10 N is needed to keep a 500 g book in place against a wall as shown.What maximum mass of book could be supported by the 10 N force if it were directed upwards at an angle of 60° instead of at right angles to the wall?
Answers
Answered by
Elena
When the force F is normal to the wall the projections of the forces acting on the book are
x: N = F,
y: mg =F(fr).
F(fr) = k•N = k•F,
k•F = m•g,
k = m•g/F =0.5•9.8/10 = 0.49.
When the force F is at 60o to horizontal the projections of the forces acting on the book are
x: N1 = F•cosα
y: m1•g =F(fr)+F•sinα
F(fr) = k•N1= k•F•cosα,
F(fr) = m1•g - F•sinα,
k•F•cosα= m1•g - F•sinα,
m1=F•( k• cosα +sinα)/g =
= 10(0.49•0.5 + 0.866)/9,8 = 1.13 kg.
x: N = F,
y: mg =F(fr).
F(fr) = k•N = k•F,
k•F = m•g,
k = m•g/F =0.5•9.8/10 = 0.49.
When the force F is at 60o to horizontal the projections of the forces acting on the book are
x: N1 = F•cosα
y: m1•g =F(fr)+F•sinα
F(fr) = k•N1= k•F•cosα,
F(fr) = m1•g - F•sinα,
k•F•cosα= m1•g - F•sinα,
m1=F•( k• cosα +sinα)/g =
= 10(0.49•0.5 + 0.866)/9,8 = 1.13 kg.
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