Engineering Mechanics The centre of percussion is below the centre of gravity of the body and is at a distance equal to h2/kG h × kG h/kG kG2/h h2/kG h × kG h/kG kG2/h ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The three forces of 100 N, 200 N and 300 N have their lines of action parallel to each other but act in the opposite directions. These forces are known as Unlike parallel forces Like parallel forces Coplanar concurrent forces Coplanar non-concurrent forces Unlike parallel forces Like parallel forces Coplanar concurrent forces Coplanar non-concurrent forces ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The law of the machine is (where P = Effort applied to lift the load, m = A constant which is equal to the slope of the line, W = Load lifted, and C = Another constant which represents the machine friction.) P = mW - C P = m/W + C P = mW + C P = C - mW P = mW - C P = m/W + C P = mW + C P = C - mW ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The moment of inertia of a solid cone of mass ‘m’ and base radius ‘r’ about its vertical axis is 2mr²/5 3mr²/5 3mr²/10 4mr²/5 2mr²/5 3mr²/5 3mr²/10 4mr²/5 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. IP) is given by(where, A = Area of the section, IG = Moment of inertia of the section about an axis passing through its C.G., and h = Distance between C.G. and the parallel axis.) IP = IG - Ah2 IP = Ah2 / IG IP = IG / Ah2 IP = IG + Ah2 IP = IG - Ah2 IP = Ah2 / IG IP = IG / Ah2 IP = IG + Ah2 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics If the body falls freely under gravity, then the gravitational acceleration is taken as -8.9 m/s2 +8.9 m/s2 +9.8 m/s2 -9.8 m/s2 -8.9 m/s2 +8.9 m/s2 +9.8 m/s2 -9.8 m/s2 ANSWER DOWNLOAD EXAMIANS APP
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