Forces(5N, 030\(^o\)), (PN, 060\(^o\)), (QN, 150\(^o\)), (3N, 180\(^o\)) and (5N, 270\(^o\)) act on a body . If the system is in quilibrium, find, correct t...
Question 1 Report
Forces(5N, 030\(^o\)), (PN, 060\(^o\)), (QN, 150\(^o\)), (3N, 180\(^o\)) and (5N, 270\(^o\)) act on a body . If the system is in quilibrium, find, correct to one decimal place, the values of P and Q
Answer Details
Measure each direction anticlockwise from the positive x-axis and resolve. For equilibrium the sum of the components in each direction is zero.
x-components:
\[5\cos30^{o} + P\cos60^{o} + Q\cos150^{o} + 3\cos180^{o} + 5\cos270^{o} = 0\] \[\tfrac{5\sqrt3}{2} + \tfrac{P}{2} - \tfrac{\sqrt3}{2}Q - 3 + 0 = 0\]y-components:
\[5\sin30^{o} + P\sin60^{o} + Q\sin150^{o} + 3\sin180^{o} + 5\sin270^{o} = 0\] \[\tfrac{5}{2} + \tfrac{\sqrt3}{2}P + \tfrac{1}{2}Q + 0 - 5 = 0\]The y-equation gives \(\sqrt3\,P + Q = 5\), so \(Q = 5 - \sqrt3\,P\).
The x-equation gives \(P - \sqrt3\,Q = 6 - 5\sqrt3\). Substituting:
\[P - \sqrt3(5 - \sqrt3 P) = 6 - 5\sqrt3 \;\Rightarrow\; 4P - 5\sqrt3 = 6 - 5\sqrt3 \;\Rightarrow\; 4P = 6\]So \(P = 1.5\), and \(Q = 5 - 1.5\sqrt3 = 5 - 2.598 = 2.402\).
\(P \approx 1.5\,\text{N}, \quad Q \approx 2.4\,\text{N}\) (to 1 d.p.).
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