angle 0is 8=arctan-(d)If the angle 0=47.4 and the radius of the circle is 1.50 m, the speed of the mass is 4.0m/s_(e)If the mass is 1.50 kg, the magnitude of the tension in the string is 217N.(f) The string breaks unexpectedly when the mass is moving exactly eastward. The cato he mass will hit the ground is 1. 8m to the point O' Solution: The second law force diagram is shown in figure (c)Apply Newton's second law of motion, we have Tcos 6-mg=0 Tsin e Solving the equations, we have 0=arctan (d=acan"→=grtn6=√98×15×tn474°=40ms (e)Tcos6-mg=0→T 1.5×98 =21.7N cOs 6 cOs 47.4 1, in which y=1.Om Then x=1 1. 8 m V9.8 I. Give the Solutions of the following problems 1. The static and kinetic coefficients of friction for a 50 kg mass on a table surface are us=0.20 and 4=0. 15. See M=50kg Figure 3. The pulley is ideal. The system is at rest in equilibrium. (a) For each mass, sketch a second law force diagram indicating schematically all forces that are acting on each mass. (b) How large can the mass m be and not move the system? What the magnitude of the frictional force on the 50 kg mass when this situation exists?(c)If m 1g is only half the maximum mass calculated in part(b), what is the magnitude of the frictional force on the 50 kg block? Indicate whether this is a force of static friction or a force kinetic frictionangle θ is gr v 2 θ = arctan .(d) If the angle θ = 47.4° and the radius of the circle is 1.50 m, the speed of the mass is 4.0m/s . (e) If the mass is 1.50 kg, the magnitude of the tension in the string is 21.7N . (f) The string breaks unexpectedly when the mass is moving exactly eastward. The location the mass will hit the ground is 1.8m to the point O´ . Solution: The second law force diagram is shown in figure. (c) Apply Newton’s second law of motion, we have ⎪ ⎩ ⎪ ⎨ ⎧ = − = r mv T T mg 2 sin cos 0 θ θ Solving the equations, we have gr v 2 θ = arctan . (d) arctan tan 9.8 1.5 tan 47.4 4.0 m/s 2 = ⇒ = = × × = o θ v gr θ gr v (e) 21.7 N cos 47.4 1.5 9.8 cos cos 0 = × − = ⇒ = = o θ θ mg T mg T (f) Using ⎪ ⎩ ⎪ ⎨ ⎧ = = 2 2 1 y gt x vt , in which y=1.0m. Then 1.8 m 9.8 2 1 4 2 = × = = × g y x v III. Give the Solutions of the Following Problems 1. The static and kinetic coefficients of friction for a 50 kg mass on a table surface are µs=0.20 and µk=0.15. See Figure 3. The pulley is ideal. The system is at rest in equilibrium. (a) For each mass, sketch a second law force diagram indicating schematically all forces that are acting on each mass. (b) How large can the mass m be and not move the system? What the magnitude of the frictional force on the 50 kg mass when this situation exists? (c) If m is only half the maximum mass calculated in part (b), what is the magnitude of the frictional force on the 50 kg block? Indicate whether this is a force of static friction or a force kinetic friction. m M=50kg Fig.3