In the simulation below, check the checkbox to show a 2-D representation of the torque due to gravity (in green) acting on the rod, and the torque due to the tension in the string (in blue) acting on the rod. The rod is hinged at the left end, and it is held in equilibrium by the string. Note that you may need to scroll the simulation window slightly (click-and-drag on the simulation) to access the check box, which is at the bottom. Also, use the buttons under the simulation to explore the three different (equivalent) methods of finding the torque because of the tension in the string. Note that the torques are measured with respect to an axis passing through the hinge. String attaches 39.80 cm from hinge Rod weight = 20.0 N String angle to vertical = -50.0 deg. Pick a method: 1 - full torque equation 2 - force components 3 - lever arm Given the limits set by the sliders in the simulation, what is the maximum magnitude possible for the torque exerted on the rod by the string, in this case? N m String attaches 40.00 cm from hinge Rod weight = 20.0 N String angle to vertical = 0.0 deg. Pick a method: 1 - full torque equation 2 - force components 3 - lever arm Note that you may get different numbers for the two parts below. As usual, it is best to solve in terms of variables first, and plug in numbers only at the end. Part (d) O Use these settings on the simulation. Weight of the rod = 12.0 N; string attaches 30.0 cm from the hinge; string angle with the vertical is -45.0. Determine the magnitude of the tension in the string. N Use these settings on the simulation. Weight of the rod = 12.0 N; string attaches 30.0 cm from the hinge; string angle with the vertical is -45.0. Determine the magnitude of the hinge force (shown in red in the simulation). NIt is often useful to treat the lower arm as a uniform rod of length L and mass M that can rotate about the elbow. Let's say you are holding your arm so your upper arm is vertical [with your elbow below your shoulder), with a 90-degree bend at the elbow so the lower arm is horizontal, as shown above. In this position, we can say that three forces act on your lower arm: the force of gravity (Mg), the force exerted by the biceps, and the force exerted at the elbow joint by the humerus (the bone in the upper arm). Let's say the biceps muscle is attached to the lower arm at a distance of UN from the elbow, moving away from the elbow toward the hand. Use g = 10 m/sz. Part (a) Let's say your lower arm has a mass of 1.30 kg. Assuming that the force the biceps muscle exerts on your lower arm is vertical, what is the magnitude of the biceps force? N Let's say your lower arm has a mass of 1.30 kg. Carrying on from part (a), determine the force that the humerus exerts on the lower arm. If this force is directed up, use a positive sign. If this force is directed down, use a negative sign. N Let's say your lower arm has a mass of 1.30 kg. You then hold a ball in your hand. The ball has the same mass as your lower arml and we will assume the ball is a distance L from the elbow. Determine the magnitude of the biceps force now. N The figure below shows four different cases involving a uniform rod of length L and mass .X is subjected to two forces of equal magnitude. The rod is free to rotate about an axis that either passes through one end of the rod, as in (a) and (b), or passes through the middle of the rod, as in (c) and (d). The axis is marked by the red and black circle, and is perpendicular to the page in each case. This is an overhead view, and we can neglect any effect of the force of gravity acting on the rod. (21) F (b) lF (c) F (d) F 6] Part (a) X X Answered - 1 attempt left A all equal B A > B = C > D C A = B > C = D D A:C>B=D E B>D>A=C F B:D>A:C m In 2022, Zdeno Chara, the captain of the Boston Bruins, retires and opens a restaurant. He mounts a sign outside. As shown in the picture, the sign is mounted on a horizontal rod. The mass of the rod is uniformly distributed along its length. The rod is connected to the side of the building by a hinge at the left end, and the rod is held horizontal by a cable that goes from the right end ofthe rod to a point H : 3.00 meters above the hinge. The sign is placed 0 = 2.00 m from the hinge. The end ofthe rod is 1.0 m beyond the sign. Part (a) Use these numbers: the weight of the rod is Mg = 42.0 N, and the weight of the sign is mg = 26.0 N. Determine the magnitude of the tension in the cable. N Use these numbers: the weight ofthe rod is Mg = 42.0 N, and the weight of the sign is mg = 26.0 N. Determine the magnitude of the vertical component of the hinge force (the force the hinge exerts on the rod). N Use these numbers: the weight of the rod is Mg = 42.0 N, and the weight of the sign is mg = 26.0 N. Determine the magnitude ofthe horizontal component of the hinge force. N Type your numeric answer and submit