a) A pivot is placed under the beam at x = 5. Calculate the total torque around the pivot and state whether the beam is rotating clockwise or counter-clockwise (b) The pivot is moved such that the beam is now in equilibrium. Find the new x-position of the pivot.
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(a) A pivot is placed under the beam at x = 5. Calculate the total torque around the pivot and state whether the beam is rotating clockwise or counter-clockwise
(b) The pivot is moved such that the beam is now in equilibrium. Find the new x-position of the pivot.
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- D1 Sign D2 A 2.40kg pole is attached to a wall with a hinge as shown. A 5.50kg sign is suspended at the other end of a pole, where D2 = 3.50m. A cable attached to the wall forms an angle a = 14.0° with the wall, and is attached to the pole a distance D = 1.60m from the hinge. Assume the origin is at the hinge. What is the torque on the pole due to the weight of the sign? 183 NmTwo forces, F1=100 N (Newton) and F2=20N, are applied on opposite sides of a door's knob at angles theta_1=30 degrees and theta_2=60 degrees respectively, causing the door to rotate (about the axis passing though its hinges). If the distance from the knob to the rotation axis is 0.8 meter, find the net torque exerted on the door and specify the direction of rotation of the door (in this problem, assume that the force F1 causes a counterclockwise rotation and F2 a clockwise rotation). A.) Net Torque = 26.14 N.m; the door rotates counterclockwise. B.) Net Torque = -25.4 N.m; the door rotates counterclockwise. C.) Net Torque = 123.8 N.m; the door rotates clockwise. D.) Net Torque = 1.784 N.m; the door rotates clockwise.A 3.0-m rod is pivoted about its left end. A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2 m from the pivot causing a ccw torque, and a force of 5.2 N is applied at the end of the rod 3.0 m from the pivot. The 5.2 N is at an angle of 25 degrees to the rod and causes a cw torque. What is the net torque about the pivot? Choose – 6.3 Nm –1.7 Nm 0.6 Nm – 0.6 Nm
- In the figure, a constant horizontal force F→app of magnitude 31.0 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 17.5 kg, its radius is 0.656 m, and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) In unit-vector notation, what is the frictional force acting on the cylinder?F y mg f.Y Frxe f,x Axis The ladder in the picture has a mass of 32 kilograms and a length 3.2 meters. What is the normal force pushing the ladder up from the floor? FN = Assume that the ladder's weight is evenly distributed, so it can be treated as a single force through the middle. If the ladder is at a 60° angle from the ground, what is the torque exerted by the weight (using the floor as the pivot point)?Force F = (-7.0 N)î + (5.0 N) ĵ acts on a particle with position vector7 = (4.0 m)î + (5.0 m) ĵ. (a) What is the torque on the particle about the origin, in unit-vector notation? = N: m (b) What is the angle between the directions of r and Torque is the cross product of a position vector (extending from a chosen point, here the origin, to the particle) and a force vector. Did you take the cross product in unit-vector notation? Do you remember how find the angle between two vectors by taking a dot product in both unit-vector notation and also in magnitude-angle notation? (You can similarly use a cross product to do this.) Do you remember how to find the magnitude of a vector from its components?
- Mutiple choice: The torque is rotational counterpart of force, but it is not force. It has tendency of rotating objects. The torque is more effective when, (1) force is larger (2) force is far from axis of rotation (3) angle between r and the force is zero a) Only (1) is correct b) Only (2) is correct c) All (1), (2), and (3) are correct d) Both (1) and (2) are correct(a) A woman opens a 1.35 m wide door by pushing on it with a force of 44.5 N applied at the center of the door, at an angle perpendicular to the door's surface. What magnitude of torque (in N · m) is applied about an axis through the hinges? (b) A girl opens the same door, using the same force, again directed perpendicular to the surface, but now the force is applied at the edge of the door. What magnitude of torque (in N · m) is applied about the axis through the hinges now?