Answer to Question #95738 in Mechanics | Relativity for stefanie

Question

An elevator car has two equal masses attached to the ceiling as shown. (Assume
m = 3.10 kg.)
a. The elevator ascends with an acceleration of magnitude 1.00 m/s2. What are the tensions in the two strings? (Enter your answers in N.)
b. The maximum tension the strings can withstand is 75.0 N. What is the maximum acceleration of the elevator so that a string does not break? (Enter the magnitude in m/s2.)

Answer 1

Solution:

Let's write Newton's second law for second mass:

$T_2-mg=ma$ ,

$T_2=m(g+a)=3.10\cdot{}(9.8+1.00)=33.48$ N.

Let's write Newton's second law for first mass:

$T_1-T_2-mg=ma$

$T_1=T_2+m(g+a)=33.48+33.48=66.96$ N.

As we can see, T₁>T₂, so to find the maximum acceleration of the elevator, we should take into account, that the tension in first string should not exceed the maximum tension the strings can withstand.

Newton's second law for first mass in this case:

$T_{1\:max}-T_{2\:max}-mg=ma_{max}$ ,

$T_{2\:max}=mg+ma_{max}$ , so

$T_{1\:max}-mg-ma_{max}-mg=ma_{max}$

$T_{1\:max}-2mg=2ma_{max}$

$a_{max}=\frac{T_{1\:max}-2mg}{2m}=\frac{T_{1\:max}}{2m}-g=\frac{75.0}{2\cdot{}3.10}-9.8=2.3$ m/s².

Answer: T₁=66.96 N, T₂=33.48 N, a_max=2.3 m/s².