The two children shown in the figure below are balanced on a seesaw of negligible mass. The first child has a mass of 25.3 kg and sits 1.44 m from the pivot. The second child has a mass of 41.7 kg and sits 0.874 m from the pivot. Use the second condition for equilibrium (net τ = 0) to calculate Fp, the supporting force (in N) exerted by the pivot. (Enter the magnitude.)

A man holds a 198-N ball in his hand, with the forearm horizontal (see the figure). He can support the ball in this position because of the flexor muscle force M, which is applied perpendicular to the forearm. The forearm weighs 18.5 N and has a center of gravity as indicated. Find (a) the magnitude of M and the (b) magnitude and (c) direction (as a positive angle counterclockwise from horizontal) of the force applied by the upper arm bone to the forearm at the elbow joint. Upper arm bone - Flexor muscle M Elbow joint cg 0.0510 mt T0.0890 m 0.330 m-

(a) A person opens a 1.20 m wide door by pushing on it with a force of 57.5 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?   _________N · m   (b) A man 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?   _________N · m

A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. (a) What is the y component of the tension in the rope? (b) What is the x component of the tension in the rope? (c) What is the magnitude of the tension in the rope? (a) y component of the tension: Ty = (b) x component of the tension: Tx = (c) What is the magnitude of the tension in the rope?

A hinged beam, 92.5 kg, is supported by a horizontal cable as in the figure. The length of the cable is 2.00 m and the length of the beam is 3.00 m. Find: (a) the torque produced by gravity on the beam, (b) the tension in the cable, (c) the horizontal and vertical components of the force from the hinge, and (d) the magnitude and direction of the force from the hinge.

The angle between the beam and the floor is 15.0 degrees. The angle between the rope and the horizontal is 31.0 degrees. The beam is 4.10-m long and has a mass of 6.10 kg. The box sits a distance of d = 0.770 m from the upper end of the beam and has a mass of 9.60 kg.What is the torque due only to the weight of the box if the axis is at the hinge?

A man holds a 195-N ball in his hand, with the forearm horizontal (see the figure). He can support the ball in this position because of the flexor muscle force M , which is applied perpendicular to the forearm. The forearm weighs 20.7 N and has a center of gravity as indicated. Find (a) the magnitude of M and the (b) magnitude and (c) direction (as a positive angle counterclockwise from horizontal) of the force applied by the upper arm bone to the forearm at the elbow joint. Upper arm bone- Flexor muscle M Elbow cg joint 0.0510 m+ 0.0890 m -0.330 m- (a) Number i Units (b) Number i Units (c) Number i Units >

I need help with problem over Rotational Kinectic (Equilibrium).. A window washer weighing 825.6 N is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 262.5 N and is 4.034 m long. What is the force in each rope when the window washer stands 1.48 m from one end? Force in Rope 1 (Rope closest to the window washer.) =  Force in Rope 2 (Rope farthest from the window washer.) =

The center of mass of the arm shown in the figure is at point A. Find the magnitudes (in N) of the tension force F and the force F which hold the arm in equilibrium. (Let 8 = 24.0°.) Assume the weight of the arm is 48.3 N. IF 8.00 cm 29.0 cm 1₂ 354 X Draw a diagram showing the forces and distances. Find the sum of the torques about point O, and set this equal to zero. Solve your equation for F₁. N 340.30 X Apply Newton's second law in the x and y directions, and solve for F Sx Pythagorean theorem to calculate F. . N and F You will need your value of F. Then use the sy'