Question

# Two blocks, each of mass $$m = 3.50 kg$$, are hung from the ceiling of an elevator as in above figure. (a) If the elevator moves with an upward acceleration $$\vec{a}$$ of magnitude $$1.60 m/s^{2}$$, find the tensions $$T_{1}$$ and $$T_{2}$$ in the upper and lower strings. (b) If the strings can withstand a maximum tension of $$85.0 N$$, what maximum acceleration can the elevator have before a string breaks.

Solution

## (a) Free-body diagrams of the two blocks are shown below. Note that each block experiences adownward gravitational force$$F_{g} = (3.50 kg)(9.80 m/s^{2} ) = 34.3 N$$Also, each has the same upward acceleration as the elevator, in this case$$a_{y} = +1.60 m/s^{2}$$.Applying Newton’s second law to the lower block:$$\sum F_{y} = ma_{y} \Rightarrow T_{2} − F_{g} = ma_{y}$$or$$T_{2} = F_{g} + ma_{y} = 34.3 N + (3.50 kg)(1.60 m/s^{2} ) = 39.9 N$$Next, applying Newton’s second law to the upper block:$$\sum F_{y} = ma_{y} \Rightarrow T_{1} −T_{2} − F_{g} = ma_{y}$$or$$T_{1} = T_{2} + F_{g} + ma_{y} = 39.9 N + 34.3 N + (3.50 kg)(1.60 m/s^{2} )$$$$= 79.8N$$(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 $$a_{y}$$ when $$T_{1} = 85.0$$. Making use of the general relationships derived in (a) above gives:$$T_{1} = T_{2} + F_{g} + ma_{y} = (F_{g} + ma_{y})+ F_{g} + ma_{y} = 2F_{g} + 2ma_{y}$$or$$a_{y}=\frac{T_{1}-2F_{g}}{2m}=\frac{85.0N-2(34.3N)}{2(3.50kg)}=2.34m/s^{2}$$.Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More