A box is 1.20 kg. The coefficient of static friction between the box and a horizontal ground is 0.250. To make the box start moving, you pull on...a string attached to it. For the string tension to...

A box is 1.20 kg. The coefficient of static friction between the box and a horizontal ground is 0.250. To make the box start moving, you pull on...

a string attached to it. For the string tension to be as small as possible, you should pull at what angle above the horizontal?

Asked on by islnds

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

Lets take the the string force is at a angle of `theta` to the horizontal.

So if we consider the situation at the moment of moving the box have to forces to the horizontal direction. One is the string force component to the horizontal direction and the other is the friction force opposite to it.

string force to rarr direction = `Tcostheta`

Friction force larr direction = `mu*R` = `0.25*1.2*9.81`

Considering Force equilibrium;

string force to rarr direction = friction force larr direction

Tcostheta = `0.25*1.2*9.81`

T = `2.943/costheta`

To T to be small as possible costheta to be maximum.

Maximum `costheta = 1`

`costheta = 1`

`theta = 0`

So to get the smallest string tension we need to pull the string parallel to the horizontal and the minimum tension will be 2.943N.

Sources:

jishnudeepkar | Student, Grade 11 | (Level 1) Honors

Posted on

sorry the thetas have not come . any ways it should be

d(costheta + sintheta/4)/dtheta=0

jishnudeepkar | Student, Grade 11 | (Level 1) Honors

Posted on

ANs = tan inverse (1/4)

jishnudeepkar | Student, Grade 11 | (Level 1) Honors

Posted on

Lets take the the string force is at a angle of   to the horizontal.

I was not satisfied with the above calculation.

The friction should be = (1.2* 9.8 - T sin)0.25

hence  Tcos= 2.94 - Tsin/4

T( cos + sin/4)= 2.94

hence for T minm bracketed quantity should be max so differentiate and equate it to zero. d(cos+ sin/4)/d=0

u probably get tan = 1/4

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