If you can label the answers that would be very helpful 1. Solve the EXAMPLE PROBLEM in the INTRODUCTION using the polygon method. Usean 8.5" x 11" sheet of paper (or the back of this page) so that you can create a large-scale vectordiagram and thereby obtain results of greater accuracy. Draw your vectors to scale using a rulerand use a protractor to establish their directions. Record your results below. Compare yourresults with the analytical results in the EXAMPLE PROBLEM. Attach your vector diagram.Magnitude of the net force%3DDirection of the net force%3D2. Use the analytical method to find the resultant of the forces F, = 25.0 N @ 37.0° andF, = 40.0 N @ 250°.%3D%3D3. Find the equilibrant of the force you found in QUESTION 2. Although Equations 4 and 5 were derived for the special case of two vectors, they can be generalisto find the magnitude and the direction angle of the resultant of any number of force vectoreFinally, the equilibrant of any number of forces is the single force required to place the obiectwhich the forces are acting in a state of static equilibrium. Such a force balances the resultant fonNote that this means that the equilibrant force is equal in magnitude to the resultant force and acte oan angle of 180° from the direction of the resultant force.EXAMPLE PROBLEMForces F = 2.50 N@ 0°, F, =one you will be using in the EXPERIMENTAL ACTIVITIES. A rectangular coordinate system haebeen added to the pictorial representation for your convenience. Find the resultant and theequilibrant of these three forces.%3D1.00 N @ 120° and F, = 1.50N@ 210° act on a ring similar to the%3D%3DyF,SOLUTIONA useful strategy in organizing the given information in problems like this one is to create a table,similar to the one shown below, that you can use in working your way methodically through thecalculations required in order to solve the problem.ForceMagnitudeX-componenty-component2.50 N(2.50 N) cos 0° = 2.50 N%3D(2.50 N) sin 0° = 0 N%3DF21.00 N(1.00 N) cos 120° = -0.50 N%3D(1.00 N) sin 120° ~ 0.866 NF,1.50 N(1.50 N) cos 210° - -1.30 N(1.50 N) sin 210° = -0.750 N%3DΣF,0.70 ΝΣF,-0.12Ν

Question

If you can label the answers that would be very helpful 1. Solve the EXAMPLE PROBLEM in the INTRODUCTION using the polygon method. Usean 8.5&#34; x 11&#34; sheet of paper (or the back of this page) so that you can create a large-scale vectordiagram and thereby obtain results of greater accuracy. Draw your vectors to scale using a rulerand use a protractor to establish their directions. Record your results below. Compare yourresults with the analytical results in the EXAMPLE PROBLEM. Attach your vector diagram.Magnitude of the net force%3DDirection of the net force%3D2. Use the analytical method to find the resultant of the forces F, = 25.0 N @ 37.0° andF, = 40.0 N @ 250°.%3D%3D3. Find the equilibrant of the force you found in QUESTION 2. Although Equations 4 and 5 were derived for the special case of two vectors, they can be generalisto find the magnitude and the direction angle of the resultant of any number of force vectoreFinally, the equilibrant of any number of forces is the single force required to place the obiectwhich the forces are acting in a state of static equilibrium. Such a force balances the resultant fonNote that this means that the equilibrant force is equal in magnitude to the resultant force and acte oan angle of 180° from the direction of the resultant force.EXAMPLE PROBLEMForces F = 2.50 N@ 0°, F, =one you will be using in the EXPERIMENTAL ACTIVITIES. A rectangular coordinate system haebeen added to the pictorial representation for your convenience. Find the resultant and theequilibrant of these three forces.%3D1.00 N @ 120° and F, = 1.50N@ 210° act on a ring similar to the%3D%3DyF,SOLUTIONA useful strategy in organizing the given information in problems like this one is to create a table,similar to the one shown below, that you can use in working your way methodically through thecalculations required in order to solve the problem.ForceMagnitudeX-componenty-component2.50 N(2.50 N) cos 0° = 2.50 N%3D(2.50 N) sin 0° = 0 N%3DF21.00 N(1.00 N) cos 120° = -0.50 N%3D(1.00 N) sin 120° ~ 0.866 NF,1.50 N(1.50 N) cos 210° - -1.30 N(1.50 N) sin 210° = -0.750 N%3DΣF,0.70 ΝΣF,-0.12Ν

Accepted Answer

To add the vectors.