Question

Two blocks each of mass m = 3.49 kg are fastened to the top of an elevator as in the figure.

(a) If the elevator has an upward acceleration a = 1.60 m/s2, find the tensions T1 and T2 in the upper and lower strings.
T1 =_____ N
T2 =_____ N

(b) If the strings can withstand a maximum tension of 94.0 N, what maximum acceleration can the elevator have before the upper string breaks?
________ m/s2

please show all work and answer

Answer 1

Given that the mass of each block is m = 3.49 kg

Acceleration of the block is a = 1.60 m/s²

(a)

      Apply Newtons second law to the lower block

                            T2 - mg = ma

                                    T2 = ma + mg

                                    T2 = m(a + g)

                                    T2 = 39.786 N

     Apply Newtons second law to the upper block

                       T1- T2 - mg = ma

                                    T1 = T2 + ma + mg

                                         = 39.786N + m(a + g)

                                         = 79.57 N

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(b) Upper string withstand a tension T = 94.0 N

                                                   

       Apply Newtons second law to the upper block  

                   T - mg - T2 = ma

                   a = (T - mg - T2) / m

                      = (94.0 N - 34.20N - 39.786N ) / 3.49 kg

                      = 5.73 m/s²                 

                                    

                                   

Answer 2

(a) Mass of each block is, m- 3.49 kg Upward acceleration of the elevator, a -1.60 m/s The two blocks experiences a downward gravitational force, that is ng= (3.49 ㎏)(98 m/s2) 34.202 N For lower block: Apply the Newton's second law of motion to the lower block, the net force on the block is ma-i-m 12= ma+mg Thus, the tension in the lower strıng 1s (3.49 kg (1.60 m/s2 +9.8 m/s) 39.786 N 39.8N (Approximately) For upper block: Apply the Newton's second law of motion to the upper block, the net force on the block is Thus, the tension in the upper string is (349kg) 1.60 m/s)+39.786 N+34.202 N 79.572 N 79.57N (Approximately) Answer 1- 79.57N and T-39.8 N From the above results, the tensi on in the upper string is greater than the tension in the lower string (7> ). So, the upper strin breaks first when the accel erati on of the system increases. Now, we find the maximum acceleration when T1 = 94.0 N The expression for the tension in the upper string is, But, 7- ma+mg, so the above equation becomes, =2ma + 2mg Rearrange the equation for a T, - 2mg 2m Thus, the maximum acceleration 1S, T - 2mg 2m 94.0 N-2 (34.202 N) 2(349 kg) 3.67 ms