2004 - JAMB Physics Past Questions and Answers - page 2

Pressure developed in the effort piston = ^F/_a

∴^F/_a = ⁴⁰/_0.4 = ⁴⁰⁰/₄ = 100Nm²

this pressure is transmitted to the larger piston of area A; => P = ^F/_A

∴ 100 = ⁴⁰⁰/_A

=> A = ⁴⁰⁰/₁₀₀ = 4m²

Coefficient of friction μ = ^F/_R = ^F/_mg = 500/1000 = 0.5

∴ μ = 0.5

Component of 8N along the horizontal = 8 cos 60^o = 8 x 0.5

= 4N

Similarity Component of 6N along the horizontal = 6 cos 60^o = 3N

∴ Total force towards the right = 10 + 4 + 3

= 17N

Again total force toward the left = 4N

∴ Net horizontal force (towards the right) = 17-4

= 13N

NOTE: Firstly, it should be noted that the mass, as well as the weight of a body is independent of its size.

∴ Let the weight of the moon = M_m x g_m

weight of the earth = M_e x g_e

But g_e = 80g_m

∴ weight of the earth = M_e x 80g_m

=> weight of the moon / weight of the earth = M_m x g_m / M_e x 80g_m

thus ^M_m/_Me ∝ ^g_m/_80gm ∝ ¹/₈₀

=> M_m:M_e ∝ 1:80

35^o =35^o = 70^o

180 - 70 = 110

110/2 = 55^o