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3 years ago

A 800N student does a handstand with both hands at an angle of 15 degrees from the vertical what is the force on each hand

Physics

1 answer:

jeyben [28]3 years ago

5 0

here we can say that net force on the student vertically upwards will be counter balance by his weight downwards

Let net force F is exerted by each hand

so here we will have

$2Fcos15 = W$

$2F cos15 = 800$

$F = \frac{800}{2 cos15}$

$F = 414.1 N$

so the force exerted by each hand will be 414.1 N

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3 years ago

Suppose you wanted to break a meter stick over your knee. The cross-section of a meter stick is rectangular. Will it be easier t

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Answer:

It will be easier to break the meter rule with the long side against my knee.

Explanation:

To break the meter rule involves the principle of bending moment. The long side will require less force to generate the same amount of bending moment that will have to be generated to break the meter rule. The short side on the other hand will require more force to generate this mount of bending moment. This is because the shorter has a very small surface area, which concentrates the force on your knee. The pressure is then dissipated as more pressure to your knee. Th longer side has a lesser surface area so, most of the force is used in breaking the meter rule.

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3 years ago

You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in

FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, $V=108\ cm^3$

We know that the volume of the cylinder is given by :

$V=\pi r^2 h$

$108=\pi r^2 h$

$h=\dfrac{108}{\pi r^2}$ ............(1)

The upper area is given by :

$A=32r^2+2\pi rh$

$A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}$

$A=32r^2+\dfrac{216}{r}$

For maximum area, differentiate above equation wrt r such that, we get :

$\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}$

$64r-\dfrac{216}{r^2}=0$

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Dividing equation (1) with r such that,

$\dfrac{h}{r}=\dfrac{108}{\pi r}$

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3 years ago

2. Three super strong teenagers pull a heavy crate across the floor. Dion pulls with a force of 18.5 N towards 0°. Shirley pulls

kogti [31]

Answer:

A) The resultant force is 43.4 [N]

B) The movement of the heavy crate is going to the right and in the negative direction on the y-axis

Explanation:

We need to make a sketch of the different forces acting on the heavy crate.

In the attached image we can see the forces and the sum of the vector with their respective angles.

Forces in the X-axis

$Fdionx=18.5N\\\\Fshix=16.5*cos(30)=14.29N\\Fjoanx=19.5*cos(60)=9.75N\\\\Forcex= 18.5 + 14.29 + 9.75 = 42.54 N$

Forces in the y-axis

$FDiony=0[N]\\Fshirley= 16.5*sin(30)=8.25[N]\\Fjoany=19.5*sin(60)=16.88 [N]\\\\Forcesy=0+8.25-16.88= -8.63[N]$

Using the Pythagorean theorem

$Tforce=\sqrt{(42.54)^{2} +(8.63)^{2} } \\\\Tforce= 43.4N$

The movement of the heavy crate is going to the right and in the negative direction on the y-axis, this can be easily seen in the graphical sum of vectors.

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3 years ago