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Design a decorative 鈥渕obile鈥 to consist of a low-mass rod of length 0.49msuspended from a string so that the rod is horizontal, with two balls hanging from the ends of the rod. At the left end of the rod hangs a ball with mass 0.484kg.At the right end of the rod hangs a ball with mass 0.273kg.You need to decide how far from the left end of the rod you should attach the string that will hold up the mobile, so that the mobile hangs motionless with the rod horizontal (鈥渆quilibrium鈥). You also need to determine the tension in the string supporting the mobile. (a) What is the tension in the string that supports the mobile? (b) How far from the left end of the rod should you attach the support string?

Short Answer

Expert verified

(a) The tension (T) in the string that supports the mobile is7.42N.

Step by step solution

01

Definition of Tension:

Tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable.

02

Define the formula of mass:

The tension(T)in the string that supports the mobile can be calculate as follows.

T=Mg 鈥.. (1)

Here,Mis the mass and gis the acceleration due to gravity having a value of 9.8m\s2.

03

Given data:

Consider the given data as below.

Mass of ball at left end,m1=0.484kg

Mass of the ball at right end,m2=0.273kg

Acceleration due to gravity,g=9.8m/s2

04

(a) The tension in the string that supports the mobile:

To hold up the mobile, you should string attach at center of mass of the system, at center of mass point total mass of the system is suppose to be concentrate.

Total mass of the system is given by

Mtotal=m1+m2=0.484kg+0.273kg=0.757kg

The tension (T)in the string that supports the mobile can be calculate as follows.

T=Mtotalg

Here, gis the acceleration due to gravity having a value of9.8m/s.

Now, substitute known values in the above equation, and you have

T=0.757kg(9.8m/s2)=7.42N

Hence, The tension in the string that supports the mobile is 7.42N.

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