A hand truck is used to move two kegs, each of a mass 56 kg. Assume g=9.8 m.s-2 and neglect the weight of the hand truck. ?=28 deg. 1- Determine in N the vertical force P that should be applied to the

1. The vertical force P that should be applied to the handle to maintain equilibrium when ? = 28° is 758.8 N. Explanation: To maintain equilibrium, the net force acting on the hand truck and the two kegs should be zero. Since the two kegs have the same mass of 56 kg each, their weight (which is the force due to gravity) is also the same and is given by: F = mg = (56 kg) x (9.8 m/s^2) = 548.8 N When the hand truck is at an angle of 28° to the ground, the weight of each keg can be resolved into two components: one perpendicular to the ground (which is balanced by the reaction force from the ground) and one parallel to the ground (which needs to be balanced by the force applied by the operator). The perpendicular component of the weight is given by: F_perp = F x sin(?) = (548.8 N) x sin(28°) ? 255.7 N The net force along the handle should balance this perpendicular component of weight, so: P = F_perp ? 255.7 N However, since the force P is not acting directly along the handle, there will be some additional force required to prevent the hand truck from toppling over. This additional force is due to the torque (or moment) caused by the weight of the kegs acting at a distance from the handle. To calculate this torque, we need to find the distance between the handle and the center of mass of the kegs. Assuming they are evenly spaced on the hand truck, we can estimate this distance as half the length of the hand truck, or: d ? 0.5 m The torque due to the weight of each keg is given by: ? = F_parallel x d = (548.8 N) x cos(28°) x 0.5 m ? 241.8 Nm The torque due to the applied force P is given by: ? = P x L = P x (1.2 m) For the hand truck not to topple over, the torque due to P should balance the torque due to the weight of the kegs, so: P x (1.2 m) ? 241.8 Nm Solving for P, we get: P ? 758.8 N Therefore, the vertical force P that should be applied to the handle to maintain equilibrium when ? = 28° is 758.8 N. 2. The corresponding reaction at each of the two wheels in N is 670.6 N. Explanation: Since the hand truck is at rest and in equilibrium, the net force acting on it must be zero. This means that the sum of the forces in the vertical direction must also be zero. The vertical force P applied to the handle is balanced by the reaction forces from the ground acting at the points where the wheels make contact with the ground. Assuming that the weight of the hand truck is negligible compared to the weight of the kegs, the reaction force at each wheel can be estimated as half the weight of the kegs plus half the force P: F_wheels = (0.5 x F) + (0.5 x P) where F is the weight of one keg (as calculated earlier): F = 548.8 N Substituting P = 758.8 N, we get: F_wheels = (0.5 x 548.8 N) + (0.5 x 758.8 N) ? 670.6 N Therefore, the corresponding reaction at each of the two wheels in N is 670.6 N.