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Question

Two blocks, each of mass m=3.50kg, are hung from the ceiling of an elevator as in above figure. (a) If the elevator moves with an upward acceleration a of magnitude 1.60m/s2, find the tensions T1 and T2 in the upper and lower strings. (b) If the strings can withstand a maximum tension of 85.0N, what maximum acceleration can the elevator have before a string breaks.
1863218_d32df97566be44f2b74a561891e0e55b.png

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Solution

(a) Free-body diagrams of the two blocks are shown below. Note that each block experiences a
downward gravitational force
Fg=(3.50kg)(9.80m/s2)=34.3N
Also, each has the same upward acceleration as the elevator, in this case
ay=+1.60m/s2.
Applying Newton’s second law to the lower block:
Fy=mayT2Fg=may
or
T2=Fg+may=34.3N+(3.50kg)(1.60m/s2)=39.9N
Next, applying Newton’s second law to the upper block:
Fy=mayT1T2Fg=may
or
T1=T2+Fg+may=39.9N+34.3N+(3.50kg)(1.60m/s2)
=79.8N
(b) Note that the tension is greater in the upper string, and this string will break first as the acceleration of the system increases. Thus, we wish to find the value of ay when T1=85.0. Making use of the general relationships derived in (a) above gives:
T1=T2+Fg+may=(Fg+may)+Fg+may=2Fg+2may
or
ay=T12Fg2m=85.0N2(34.3N)2(3.50kg)=2.34m/s2.
1801687_1863218_ans_5ae579e46493422f87d3ae7ca6820b5c.png

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