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Greeley [361]
3 years ago
9

A gymnast with mass 46.0 kg stands on the end of a uniform balance beam as shown. The beam is 5.00 m long and has a mass of 250

kg (excluding the mass of the two supports). Each support is 0.540 m from its end of the beam. What are the forces on the beam due to support 1 (F1) and support 2 (F2)?
Physics
1 answer:
worty [1.4K]3 years ago
3 0
Sum up the moments of the right and the left support:
∑ M2 = 0
F1 · 3.92 m = 46 kg · 9,81 m/s² · 4.46 m + 250 kg · 9.81 m/s² · 1.96 m
F 1 · 3.92 = 2012.6 + 4806.9
F 1 = 6819.5 : 3.92
F 1 = 1739.67 N
∑ M 1 = 0
F 2 · 3.92 m + 46 kg · 9.81 m/s² · 0.54 m = 250 kg · 9.81 m/s² · 1.96 m
F 2 · 3.92 + 243.68 = 4806.9
F 2 · 3.92 = 4563.22
F 2 = 4563.22 : 3.92
F 2 = 1164 N
Answer: The forces on the beam are: F 1 = 1739.67 N and F 2 = 1164 N. 
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A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/
hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is 9.1\cdot 10^{-4} J

C) The work done by the electric field is -2.4\cdot 10^{-3} J

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the particle

\theta is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

  • The force is directed vertically upward (because the field is directed vertically upward)
  • The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

\theta=90^{\circ}

and cos 90^{\circ}=0: therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

\theta=0^{\circ}

and

cos 0^{\circ}=1

Therefore, the work done by the electric force is

W=Fd

and we have:

F=qE is the magnitude of the electric force. Since

E=4.30\cdot 10^4 V/m is the magnitude of the electric field

q=32.0 nC = 32.0\cdot 10^{-9}C is the charge

The electric force is

F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

\theta=90^{\circ}+45^{\circ}=135^{\circ}

Moreover, we have:

F=1.38\cdot 10^{-3} N (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

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A helium atom (mass 4.0 u) moving at 598 m/s to the right collides with an oxygen molecule (mass 32 u) moving in the same direct
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Given that initially both are moving in the same direction and lets take they are moving in the right direction.

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598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445

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