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 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

Answer 1

As options are given, we'll go for an indirect approach.

Tsinx TcoSx 50° Vmg 500 VMg

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) = 40.5⁰

Option d)   (mg +Mg)/T = 3577/3997

=0.89

x = sin^-1(0.89) = 62.9⁰

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 =40.5⁰ .

So the value of T corresponding to x =40.5⁰ is option c, T = 5476 N