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13. A force of 20 N at angle of 30^circ to the horizontal and a force F_(2) at an angle of 60^circ to the horizontal are applied on an object as shown in the Figure below so as to make the object in equilibrium. Calculate the magnitude of the force F_(2) and weight of the object. square

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13. A force of 20 N at angle of 30^circ to the horizontal and a force F_(2) at an angle of 60^circ to the horizontal are applied on an object as shown in the Figure below so as to make the object in equilibrium. Calculate the magnitude of the force F_(2) and weight of the object. square

13. A force of 20 N at angle of 30^circ to the horizontal and a force F_(2) at an angle of 60^circ to the horizontal are applied on an object as shown in the Figure below so as to make the object in equilibrium. Calculate the magnitude of the force F_(2) and weight of the object. square

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GideonAdvanced · Tutor for 1 years

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To solve this problem, we will use the principles of static equilibrium, which state that the sum of all forces in any direction must be zero for an object to be in equilibrium. We have two forces, \(F_1\) and \(F_2\), acting at angles to the horizontal, and the weight \(W\) acting vertically downward.Given:- \(F_1 = 20\) N at an angle of \(30^\circ\) to the horizontal- \(F_2\) is unknown, at an angle of \(60^\circ\) to the horizontal- The weight \(W\) is also unknownWe will resolve the forces into their horizontal (x-axis) and vertical (y-axis) components and then apply the equilibrium conditions.Step 1: Resolve \(F_1\) into horizontal and vertical components.- The horizontal component of \(F_1\) is \(F_{1x} = F_1 \cos(30^\circ)\)- The vertical component of \(F_1\) is \(F_{1y} = F_1 \sin(30^\circ)\)Step 2: Resolve \(F_2\) into horizontal and vertical components.- The horizontal component of \(F_2\) is \(F_{2x} = F_2 \cos(60^\circ)\)- The vertical component of \(F_2\) is \(F_{2y} = F_2 \sin(60^\circ)\)Step 3: Apply the equilibrium condition for the horizontal direction.Since the object is in equilibrium, the sum of the horizontal forces must be zero:\(F_{1x} - F_{2x} = 0\)Step 4: Apply the equilibrium condition for the vertical direction.The sum of the vertical forces must also be zero:\(F_{1y} + F_{2y} - W = 0\)Now let's calculate the components and solve for \(F_2\) and \(W\).\(F_{1x} = 20 \cos(30^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\) N\(F_{1y} = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10\) NFrom Step 3, we have:\(10\sqrt{3} - F_2 \cos(60^\circ) = 0\)\(F_2 \cos(60^\circ) = 10\sqrt{3}\)Since \(\cos(60^\circ) = \frac{1}{2}\), we can solve for \(F_2\):\(F_2 \times \frac{1}{2} = 10\sqrt{3}\)\(F_2 = 20\sqrt{3}\) NFrom Step 4, we have:\(10 + F_2 \sin(60^\circ) - W = 0\)Substitute \(F_2\) into the equation:\(10 + (20\sqrt{3}) \sin(60^\circ) - W = 0\)Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we can solve for \(W\):\(10 + 20\sqrt{3} \times \frac{\sqrt{3}}{2} - W = 0\)\(10 + 30 - W = 0\)\(W = 40\) NFinal Answer:- The magnitude of the force \(F_2\) is \(20\sqrt{3}\) N.- The weight of the object \(W\) is \(40\) N.
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