Question

A horizontal bar 2.421 m long with a mass of 87.71 kg is hinged to a wall. The bar is supported from its other end by a cable attached to the wall. The cable has a tension of 850.4 N. What is the angle between the cable and the bar? (in deg) A: 7.967 B: 9.958 C: 1.245 x 10^1 D: 1.556 x 10^1 E: 1.945 x 10^1 F: 2.431 x 10^1 G: 3.039 x 10^1 H: 3.799 x 10^1

          A horizontal bar 2.421 m long with a mass of 87.71 kg is hinged to a wall. The bar is supported from its other end by a cable attached to the wall. The cable has a tension of 850.4 N. What is the angle between the cable and the bar?
(in deg)
A: 7.967
B: 9.958
C: 1.245 x 10^1
D: 1.556 x 10^1
E: 1.945 x 10^1
F: 2.431 x 10^1
G: 3.039 x 10^1
H: 3.799 x 10^1

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A horizontal bar 2.421 m long with a mass of 87.71 kg is hinged to a wall. The bar is supported from its other end by a cable attached to the wall. The cable has a tension of 850.4 N. What is the angle between the cable and the bar?
(in deg)
A: 7.967
B: 9.958
C: 1.245 x 10^1
D: 1.556 x 10^1
E: 1.945 x 10^1
F: 2.431 x 10^1
G: 3.039 x 10^1
H: 3.799 x 10^1

Added by Mar-A M.

University Physics with Modern Physics

Wolfgang Bauer, Gary D. Westfall 2nd Edition

Chapter 11

Instant Answer

Solved by Expert Adi S

06/14/2023

Step 1

We can do this by multiplying the mass of the bar by the acceleration due to gravity (9.81 m/s^2): F_weight = m*g F_weight = 87.71 kg * 9.81 m/s^2 F_weight = 859.1 N Show more…

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A horizontal bar 2.421 m long with a mass of 87.71 kg is hinged to a wall. The bar is supported from its other end by a cable attached to the wall. The cable has a tension of 850.4 N. What is the angle between the cable and the bar? (in deg) A: 7.967 B: 9.958 C: 1.245x10^1 D: 1.556x10^1 E: 1.945x10^1 F: 2.431x10^1 G: 3.039x10^1 H: 3.799x10^1

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Transcript

00:01 So, let us start with the concept which we are going to use here for this question.

00:06 Since we know the sum of the force, this is going to be equals to zero according to newton's second law.

00:13 So, according to question let's suppose this is nothing but a wall by which a rod is supported in such a way that by an, what we will say, a cable, okay, and a load is applied onto this one end of the rod.

00:32 So, length of the rod we have been given 2 .421 meters.

00:39 The mass of m is equals to 87 .71 of kg.

00:44 So, in that case, whatever the tension that is exerted on the cable, that we have been given 850 .4 of newtons.

00:54 So, this angle, if we suppose as theta, so this angle theta is basically we have been asked to find out.

01:03 So, according to newton's law, since we know the summation of the force is equals to zero, either in the horizontal direction or in the vertical direction.

01:14 So, basically this tension would have the two components.

01:18 One will be the vertical one and one will be the horizontal.

01:21 So, if i am equating the vertical components here, so vertical components are nothing but mg times of this l by 2 minus of t sine of theta into l would be equals to zero.

01:38 How this comes? because the mass will exert in downward direction as mg, okay, and this is nothing but at a distance of what? l by 2.

01:47 That is why we have taken l by 2...

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