my professor didn't use formula steps on all these problems but he only used answers...so please look at those all questions and fill the formula steps on each of those questions....i just need formula steps

3. The sail of a sailboat generates a lift force of 10,000 NV as shown in the diagram. The force is concentrated at point A. For the geometry shown, determine the torque experienced at the rudder, point B. Clearly provide your position vector and the 0-50 force components. If there is no correction made to the rudder, based on your torque components, which direction, CW or CCW, will the boat rotate? A 10,000N The force is: 12 m F = (10000 sin 50 , 10000 cos 50,0) ~ (7660.4, 6427.9,0) N B The position vector is:' r = (4, 12,0) m The torque vector is then: T= rx F = (0 , 0, -66213.2) N-m A negative sign means a clockwise turning moment. So, the boat will be rotating clockwise. That implies you will need to provide some rudder change to keep the boat from turning.5. The rotor of a tilt-rotor aircraft is currently making an angle 0 = 40 as shown in the diagram. Each rotor generates a force of F = 4500/bs. . The rotor is located 20 ft behind the nose and 15 ft above the nose. Place the axis origin at the nose of the aircraft as shown in the diagram. Find the torque applied at the N509PA aircraft nose. Clearly provide your position vector and the force components AW609 For the force vector: F = (-4500 cos 40 , 4500 sin 40 , 0) = (-3447.2 , 2892.5 , 0) lbs For the position vector: r = (20 , 15, 0) ft In actual engineering analysis, the vertical axis would have been the z-axis. So that the The torque vector is thus: T = rx F =(0 , 0, 109558) ft-lbs position vector would have been (20, 0, 15). The same thing also applies to the force (0 , 0, 219116) ft-lbs components. The torque vector would have There are 2 rotors, so the total torque is: been (0, 219116, 0) denoting a rotation about the y-axis, also known as the pitch-axis. But let's not worry about those in this class.3. A box has a weight of 80 lbs. is being suspended by 2 ropes AC and AB as shown. The Rope AB is to remain in a horizontal position and Rope AC makes an angle Q= 320 . What are the forces along the Ropes AB, AC if the box is to stay stationary. Provide both the magnitudes and the components of the forces. B This is an application of force components. Denote the forces in the 2 ropes as: AC, AB . For the box to stay put, the vertical and horizontal force components must balance each other out, in engineering language: the sums of the forces must be zero. You now have: AC.= -|AC|cos 320 i + ||AC|sin 320 j AB = |AB| i+ 0j Note that the rope AB has no vertical component AC sin 320 =80 The vertical component must support the weight, so that: AC ~ 151lbs |AC/ cos 320 - JAB . The horizontal components must balance each other out: AB ~ 128.1lbs.Automatic Zoom 6. This is a continuation of the problem from Quiz 1: A tractor pulls down a dead tree with a force F = 2000 N . The rope is attached to the tree at point A and the position of the tractor is located at point B. The origin of the 3-D axis is chosen to be the base of the tree, point 0. Determine the torque applied at point A. From the diagram, you see that A( 0,0,12) and B(4,12,0) 12 m 2000 N Therefore: AB = 47 +12j-12k and HAB 127 124 V304 V304 304 UAB The unit vector defines the direction for the force, so the force components are 12 m simply scaled by this unit vector: Am 4i F =2000 12 j 12k V304 304 1304 F ~ (458.87 +1376.5j-1376.5k) The diagram reveals that OA = 0/ +0/ +12k The torque vector is then: T = OA XF = (0,0,12)x (458.8,1376.5,-1376.5) 7 ~ (-16518, 5505.6 , 0) N-m |7-17411.4 N-m The signs of the torque vector means that: Along the x-axis, there is a clockwise turning moment; along the y- axis, there is a counter-clockwise turning moment, both working together to bring down the tree.2. Find the tensions in Cables 1 and 2 to keep the 100-kg block from moving. Provide both the magnitudes and the components of the 550 30 tensions. Note: While it is common knowledge but just to make sure that everyone is aware, the weight of the block is W = mg Cable I Cable 2 Using g = 9.8 m/sec , the block weighs 980/. After this, the 100 kg problem is virtually identical to the one we went over in class. 71 ~ 852N The tensions are: 564N In component form: T ~(-489 , 698) N and T, ~(488 , 282 N8. A vector in 3-D space has 2 known direction angles: o - 480 and B - 620. Find the missing direction angle y = ?' The direction cosines has the property: cos a + cos B + cos y =1. You already know a, B so you can find y cos y = 0.3319 7 54.8 cosy = +v0.3319 - 125.29 Note that there are 2 possible angles because when you solve a trig equation, you may have 2 solutions. In addition, you do not know exactly the specific components of this vector so you must account for 2 possible directions.You can also construct the axis system as shown here. Note that the x-axis is chosen so that it is perpendicular to the applied force. With this axis system: The position vector is: 0.16 ft 60 r = (0.16 sin 60 , 0.16 cos 60 , 0) ft 2000-16 X The force is: F = (0, -2000, 0) 1bs The torque vector is thus: T = rx F = (0 , 0, -277.1) ft- Ibs This choice of axis system gives you different position vector and the force vector but the torque vector is the same. The intent of this problem is to show you the choice of the axes is actually a decision that you need to take. With this being a math class, we do not really emphasize this topic. But where to put your axis is a major topic (and a challenge) in your future studies.Name: 4. A swimmer is attempting to head directly across a river (indicated by the dashed arrow). He swims at a velocity Vs = 1.5 mph . The river current is moving at a velocity Vc =0.8 mph as shown in the diagram. What angle 0 that he must aim if he wants to continue heading directly across the Vs = 1.5 A river? Vc = 0.8 V s has the horizontal and vertical components as: (-1.5 sin 0, 1.5 cose) V c has the horizontal and vertical components as: (0.8, 0) To head directly across, he must cancel out the current velocity. This means: To cancel the river cross current: -1.5sin0+0.8=0. You can find the angle 0 from this. 0 ~ 32.20