### Diagram Explanation:The image shows a physical setup involving a wall, a horizontal rod, a plant, and a supporting cable. Here's a breakdown of the components and forces illustrated:- **Rod:** Horizontally positioned, with a mass of 7.8 kg. The rod’s weight is considered to be concentrated at its center of mass, which is located near the midpoint. - **Plant:** Hanging vertically from the end of the rod, with a mass of 30 kg, positioned 1.8 meters from the wall.- **Cable:** Supports the rod at an angle of 38°, applying tension (T) to help maintain equilibrium.- **Forces:** - **\( T \):** Red arrow indicating the tension in the cable, directed along the cable. - **\( F_y \) and \( F_x \):** Blue arrows at the wall represent the vertical and horizontal components of the reaction force at the wall, respectively.- The black and white circle represents the pivot point or point of rotation at the wall.### Problem Description:The task involves calculating the torque acting on the system:- **Given Data:** - Plant mass: 30 kg - Distance of the plant from the wall: 1.8 meters - Rod mass: 7.8 kg - Rod length: 2 meters - Cable angle: 38°- **Objective:** Determine the magnitude of the downward torque exerted by both the rod's weight and the weight of the plant, using the wall as the axis of rotation. **Problem Statement:**The plant in the picture has a mass of 30 kg and is hanging at a distance of 1.8 meters from the wall. The horizontal rod has a mass of 7.8 kg. Assume that its weight is evenly distributed, therefore it can be treated as a single force at the center of mass. The rod is 2 meters long, and there is a cable at a 38° angle supporting it at the end.**Task 1: Torque Calculation**Using the wall as the axis of rotation, find the magnitude of the downward torque, from both the weight of the rod and the weight of the plant.\[\tau = \_\_\_\_\_ \, \text{N} \cdot \text{m}\]**Task 2: Tension Force Calculation**The downward torque is balanced by the upward torque from the force of tension. Find the magnitude of the force of tension.\[T = \_\_\_\_\_ \, \text{N}\]**Task 3: Horizontal Force Calculation**\( F \) is the contact force between the rod and the wall. Using the other horizontal force in the problem, find the horizontal component of \( F \) (the normal force) that must be present for the rod to be at equilibrium.\[F_x = \_\_\_\_\_ \, \text{N}\]**Task 4: Vertical Force Calculation**There is a vertical force from a component of the tension, but this is not enough to balance the downward forces. Find the vertical force that must be present for the wall to keep the rod up.\[F_y = \_\_\_\_\_ \, \text{N}\]

Given mass of the rod is M = 7.8 kg mass of the plant is m = 30 kg suspended at x = 1.8 m from the…

The plant in the picture has mass of 30 kg, and is hanging at a distance of 1.8 meters from the wall. The horizontal rod has mass of 7.8 kg. Assume that its weight is evenly distributed, therefore it can be treated as a single force at the center of mass. The rod is 2 meters long, and there is a cable at a 38° angle supporting it at the end. Using the wall as the axis of rotation, find the magnitude of the downward torque, from both the weight of the rod and the weight of the plant. N.m The downward torque is balanced by the upward torque from the force of tension. Find the magnitude of the force of tension. T = F is the contact force between the rod and the wall. Using the other horizontal force in the problem, find the horizontal component of F (the normal force) that must be present for the rod to be at equilibrium. N. %3D There is a vertical force from a component of the tension, but this is not enough to balance the donward forces. Find the vertical force that must be present for the wall to keep the rod up. Fy = %3D

The plant in the picture has mass of 30 kg, and is hanging at a distance of 1.8 meters from the wall. The horizontal rod has mass of 7.8 kg. Assume that its weight is evenly distributed, therefore it can be treated as a single force at the center of mass. The rod is 2 meters long, and there is a cable at a 38° angle supporting it at the end. Using the wall as the axis of rotation, find the magnitude of the downward torque, from both the weight of the rod and the weight of the plant. N.m The downward torque is balanced by the upward torque from the force of tension. Find the magnitude of the force of tension. T = F is the contact force between the rod and the wall. Using the other horizontal force in the problem, find the horizontal component of F (the normal force) that must be present for the rod to be at equilibrium. N. %3D There is a vertical force from a component of the tension, but this is not enough to balance the donward forces. Find the vertical force that must be present for the wall to keep the rod up. Fy = %3D

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

To study torque experimentally, you apply a force to a beam. One end of the beam is attached to a pivot that allows the beam to rotate freely. The pivot is taken as the origin or your coordinate system. You apply a force of F = Fx i + Fy j + Fz k at a point r = rx i + ry j + rz k on the beam. Part (a) Enter a vector expression for the resulting torque, in terms of the unit vectors i, j, k and the components of F and r. Part (b) Calculate the magnitude of the torque, in newton meters, when the components of the position and force vectors have the values rx = 4.42 m, ry = 0.015 m, rz = 0.035 m, Fx = -4.2 N, Fy = 6.6 N, Fz = 2.8 N. Part (c) If the moment of inertia of the beam with respect to the pivot is I = 368 kg˙m2, calculate the magnitude of the angular acceleration of the beam about the pivot, in radians per second squared.
While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contracting the muscles in the back of the upper leg. The weight has mass 13.5 kg, and is a distance of 28.0 cm from the knee joint. The moment of inertia of the lower leg is 0.900 kg m², the muscle force is 1470 N, and its effective perpendicular lever arm is 3.55 cm. Find the magnitude of the angular acceleration a of the weight. a = rad/s? Fow much work W is done if the leg rotates through an angle of 20.0° with a constant force exerted by the muscle? W = JI
One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 38.9° angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned. d= i >
A solid cylinder is mounted above the ground with its axis of rotation oriented horizontally. A rope is wound around the cylinder and its free end is attached to a block of mass 57.5 kg that rests on a platform. The cylinder has a mass of 225 kg and a radius of 0.480 m. Assume that the cylinder can rotate about its axis without any friction and the rope is of negligible mass. The platform is suddenly removed from under the block. The block falls down toward the ground and as it does so, it causes the rope to unwind and the cylinder to rotate. (a) What is the angular acceleration, in rad/s2, of the cylinder? rad/s2 (b) How many revolutions does the cylinder make in 5 s? rev (c) How much of the rope, in meters, unwinds in this time interval? m
The figure shows a jet engine suspended beneath the wing of an airplane. The weight of the engine is 11900 N and acts as shown in the figure. In flight the engine produces a thrust of 68600 N that is parallel to the ground. The rotational axis in the figure is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust.
John is pushing his daughter Rachel in a wheelbarrow when it is stopped by a brick 8.00 cm high (see the figure below). The handles make an angle of 0 = 20.0° with the ground. Due to the weight of Rachel and the wheelbarrow, a downward force of 407 N is exerted at the center of the wheel, which has a radius of 18.0 cm. Assume the brick remains fixed and does not slide along the ground. Also assume the force applied by John is directed exactly toward the center of the wheel. (Choose the positive x-axis to be pointing to the right.) & (a) What force (in N) must John apply along the handles to just start the wheel over the brick? 1690.53 X Your response differs from the correct answer by more than 10%. Double check your calculations. N (b) What is the force (magnitude in kN and direction in degrees clockwise from the -x-axis) that the brick exerts on the wheel just as the wheel begins to lift over the brick? magnitude KN direction ° clockwise from the -x-axis
The figure below shows a beam of uniform density with a mass of 33.0 kg and a length l = 4.45 m. It is suspended from a cord at a distance d = 1.20 m from its left end, while its right end is supported by a vertical column. (a) What is the tension (in N) in the cord? N (b) What is the magnitude of the force (in N) that the column exerts on the right end of the beam? N Need Help? Read It
A 10.0-kg monkey climbs a uniform ladder with weight w = 1.09 x 10 N and length L = 3.40 m as shown in the figure below. The ladder rests against the wall at an angle of 6 = 60°. The upper and lower ends of the ladder rest on frictionless surfaces, with the lower end fastened to the wall by a horizontal rope that is frayed and that can support a maximum tension of only 80.0 N. Rope (a) Draw a force diagram for the ladder. (Submit a file with a maximum size of 1 MB.) (b) Find the normal force exerted on the bottom of the ladder. N (c) Find the tension in the rope when the monkey is two-thirds of the way up the ladder. N (d) Find the maximum distance that the monkey can climb up the ladder before the rope breaks. m (e) If the horizontal surface were rough and the rope were removed, how would your analysis of the problem be changed and what other information would you need to answer Parts (c) and (d)?
A 740 N bear is standing on a metal plank supported by the cable shown in the figure. At the end of the plank hangs a basket weighing 80 N. Assume the plank is uniform, weighs 200 N, is 5.50 m long, and 0 = 60.0°. Goodies (a) When the bear is at x = 1.10 m, find the tension in the cable supporting the plank and the components of the force exerted by the wall on the left end of the plank. (Enter the magnitudes of your answers in N.) T = F = F = N N N (b) If the cable can withstand a maximum tension of 875 N, what is the maximum distance (in m) the bear can walk before the cable breaks? (Measure this distance from the wall.) m
Question 4. A rod is lying on the top of a table. One end of the rod is hinged to the table so that the rod can rotate freely on the table top. Two forces, both parallel to the table top, act on the rod at the same place. One force is directed perpendicular to the rod and has a magnitude of 31 N. The second force has a magnitude of 52 N and is directed at an angle with respect to the rod. If the sum of the torques due to the two forces is zero, what must be the angle? Ans: 36.59°
There is this ladder with a length L = 3.5 m and mass M= 15 kg leans against a smooth vertical wall, while its bottom legs rest on a rough horizontal floor. There is also static friction between floor and ladder is u = 0.47. The ladder makes an angle 0 = 55° with respect to the floor. A person of mass 8M stands on the ladder a distance d from its base. What is the magnitude of the normal force N, in newtons, exerted by the floor on the ladder? What is the largest distance up the ladder dmax, in meters, that the person can stand without the ladder slipping?
When a gymnast weighing 740 N executes the iron cross as in figure (a), the primary muscles involved in supporting this position are the latissimus dorsi ("lats") and the pectoralis major ("pecs"). The rings exert an upward force on the arms and support the weight of the gymnast. The force exerted by the shoulder joint on the arm is labeled F while the two muscles exert a total force F on the arm. Determine the magnitude of the force F Note that one ring supports half the weight of the gymnast, which is w 370 N as indicated in figure (b). Assume that the force F acts at an angle of 45° below the horizontal at a distance of 4.0 cm from the shoulder joint. In your estimate, take the distance from the shoulder joint to the hand to be L = 75 cm and ignore the weight of the arm. m kN Shoulder joint 4.0 cm- 45.00 Ed Bock/CORBIS