What is Mechanical Advantage (M.A.) in the principle of moments for a lever? Find the reactions forces at knife edge as shown in the figure:
Where AB is a metal bar of $70cm$ length having mass $4kg$ and a $6kg$ load is suspended from $30cm$ from the end A.
Where AB is a metal bar of $70cm$ length having mass $4kg$ and a $6kg$ load is suspended from $30cm$ from the end A.
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Mechanical Advantage (M.A.) in the principle of moments $\frac{\text{Load}}{\text{Effort}}=\frac{\text{Effort arm}}{\text{Load arm}}$
Figure shows the rod AB, the positions of the knife edges $K_1$ and $K_2$ the centre of gravity of the rod at G and the subtended weight at P. Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross-section and homogeneous; hence G is at the centre of the rod;AB = 70cm, AG = 35cm, AP = 30cm, PG = 5cm,
$AK_1, = BK_2 = 10cm$ and $K, G = K_1G = 25cm$
Also, W = Weight of the rod = 4gN and W, + S (Suspended weight) = gN. $R_1$, and $R_2$, are the normal reactions of the support at the knife edges.
For translational equilibrium of the rod.
$R_1, + R_2 - W_1- W = 0$
$\Rightarrow R_1, + R_2 - 4g - 6g = 0$
or $R_1, + R_2 = 98N ...(1)$
$W_1$, and W act vertically down and $R_1$, and $R_2$, act vertically up.
For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments about is G. The moments of $R_2$ and $W_1$ are anticlockwise (+ve), whereas the moment of $R_1$ is clockwise (-ve).
For rotational equilibrium,
$-R_1(K_1G) + W_1(PG) + R_2(K_2G) = 0 ...(2)$
Given, $W = 4gN$ and$ W_1 = 6gN$, where g = acceleration
due to gravity. We take $g = 9.8ms^{-2}$
From (2) $- 0.25R, + 0.05W_1 + 0.25R_2 = 0$
$\Rightarrow R_1 - R_2 = 1.2gN = 11.76N ...(3)$
From (1) and (3), $R_1 = 54.88N, R_2 = 43.12N$ Thus the reactions of the support are about 55N at $K_1$ and 43N at $K_2$.
Figure shows the rod AB, the positions of the knife edges $K_1$ and $K_2$ the centre of gravity of the rod at G and the subtended weight at P. Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross-section and homogeneous; hence G is at the centre of the rod;AB = 70cm, AG = 35cm, AP = 30cm, PG = 5cm,
$AK_1, = BK_2 = 10cm$ and $K, G = K_1G = 25cm$
Also, W = Weight of the rod = 4gN and W, + S (Suspended weight) = gN. $R_1$, and $R_2$, are the normal reactions of the support at the knife edges.
For translational equilibrium of the rod.
$R_1, + R_2 - W_1- W = 0$
$\Rightarrow R_1, + R_2 - 4g - 6g = 0$
or $R_1, + R_2 = 98N ...(1)$
$W_1$, and W act vertically down and $R_1$, and $R_2$, act vertically up.
For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments about is G. The moments of $R_2$ and $W_1$ are anticlockwise (+ve), whereas the moment of $R_1$ is clockwise (-ve).
For rotational equilibrium,
$-R_1(K_1G) + W_1(PG) + R_2(K_2G) = 0 ...(2)$
Given, $W = 4gN$ and$ W_1 = 6gN$, where g = acceleration
due to gravity. We take $g = 9.8ms^{-2}$
From (2) $- 0.25R, + 0.05W_1 + 0.25R_2 = 0$
$\Rightarrow R_1 - R_2 = 1.2gN = 11.76N ...(3)$
From (1) and (3), $R_1 = 54.88N, R_2 = 43.12N$ Thus the reactions of the support are about 55N at $K_1$ and 43N at $K_2$.
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