### The Circled Answer Is Not Necessarily The Correct

##### Question

# The circled answer is not necessarily the correct answer un A 280.0 kg uniform beam is supported by a cable. The beam is pivoted at the bottom, and a 85.00 kg mass hangs from its top (see fig…

un A 280.0 kg uniform beam is supported by a cable. The beam is pivoted at the bottom, and a 85.00 kg mass hangs from its top (see figure). Find the tension in the cable if the angle 50. a. 3354 N b. 1689 N c. 5476 N d. 3997 N 1467 N height

## Solutions

##### Expert Solution

As options are given, we'll go

for an indirect approach.

Balancing all the vertical forces,

Tsinx = mg + Mg

sinx = (mg+Mg)/T

So, x = sin^{-1}((mg + Mg)/T)

As the value of sine ranges from 0 to 1, (mg+Mg)/T

should be less than or equal to 1

mg +Mg = (280+85)*9.8

= 3577 N

From the given options substitute the various values of T

Option a) (mg +Mg)/T = 3577/3354

=1.066

Option b) (mg +Mg)/T = 3577/1689

= 2.117

Option c) (mg +Mg)/T = 3577/5476

=0.65

x = sin^{-1}(0.65) = 40.5^{0}

Option d) (mg +Mg)/T = 3577/3997

=0.89

x = sin^{-1}(0.89) = 62.9^{0}

Option e) (mg +Mg)/T = 3577/1467

=2.4

Options a,b, e are greater than one. So the values of T can be

neglected. From the diagram, we can see that x is an acute angle.

Therefore x =40.5^{0} .

So the value of T corresponding to x =40.5^{0} is option

c, **T = 5476 N**