PHYS 218 Chapter 5

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Condition of Equilibrium

In a system moving with a constant velocity (acceleration = 0), the sum of all the forces should be 0

Can be applied to several objects within a system:

Example

Two ropes are attached to an object of mass $m$ and then to the wall. One rope forms an angle of 45° and the other an angle of 30°

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Monday, February 21, 2011

$\sum F=m_{obj}{a}_{obj}$

Equilibrium Example

A rope of mass $m_R$ is lifting an object of mass $m_L$ at a constant velocity.

Forces:

• On object:
  • Tension of rope on object $T_{RonL}$
  • Weight of object $w_L$
• On rope:
  • Tension of object on rope $T_{LonR}$
  • Weight of rope $w_R$
  • Tension of ceiling/pulley on rope $T_{ConR}$
• On ceiling:
  • Tension of Rope on ceiling $T_{RonC}$

Static Friction Example

Find the magnitude of the force such that a block pressed to a wall does not drop.

Given:

• Force applied = $F$
• Weight = $w=mg$
• Static Coefficient = ${\mu }_S$

Inferred:

• Normal force = $N$
• Static friction = $f_s$

Initial Equations

• $x:N-F=0\therefore N=F$
• $y:f_s-w=0\therefore f_s=mg$

Solution:

• $f=mg\le {\mu }_{S}N={\mu }_{S}F$
• $F\ge \frac{mg}{{\mu }_S}$

Example from Book

• Block A weighs 1.20 N $w_A$
• Block B weighs 3.60 N $w_B$
• Coefficient between all surfaces = 0.3 ${\mu }_K$
• What force F is necessary to drag block B if A is held static by tension $T$

Surface between A & B

Normal: $N_1$

Movement: $f_1={\mu }_K{N}_1$

Surface between B & Ground

Normal: $N_2$

Movement: $f_2={\mu }_K{N}_2$

Sum of forces on Block B

$\sum F=m_B{a}_B$

Sum of forces on Block A

$\sum F=m_A{a}_A$

Forces for block A

$F_1=T={\mu }_K{\mathbb{\mathbb{N}}}_1;F_1={\mu }_K{w}_A$

$N_1=w_A$

Forces for Block B

$N_2-N_1-w_B=0;N_2=w_b+w_a$

$F=F_1+F_2={\mu }_K{w}_A+{\mu }_K(w_A+w_B)=1.8N$

Apparent Weight

Scales measure the force that you exert on them, which is equivalent to the force that the scale (ground) applies on you, which in equilibrium, is equal to the force of your weight.

Person on a scale in an elevator

elevator moving up with acceleration of 2 m/s²

Free-body diagrams remain the same: the only forces acting on the person are the weight and the Normal:

$N-w=maa=2,w=mg$

The scale calculates your mass by dividing the Normal force by an expected gravity = 9.8 m/s².

$\frac{N}{g}=m_{apparent}=m\left(\frac{a+g}{g}\right)\approx 1.2m$

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Monday, February 28, 2011

Forces in Uniform Circular Motion

Use $\hat{r},\hat{\varphi }$ reference frame, where r is direction of radius and φ is tangent to circle.

$\{\begin{array}{ll}\hat{r}& \sum F_r=m{a}_r=-m\frac{v^2}{R}\\ \hat{\varphi }& \sum F_{\varphi }=m{a}_{\varphi }=0\end{array}$

Example

Object:

• unknown mass (but it has mass)
• moving in a cone of aperture angle α in uniform circular motion of radius R.

All in presence of gravity g.

Find the period of revolution.

coordinate system:

• $r$ along radius to right
• $\varphi$ along circle (into page)
• $z$ up and down

$\{\begin{array}{ll}\hat{z}& -mg+N\sin \alpha =0\\ \hat{r}& -N\cos \alpha =-m\frac{v^2}{R}=m\frac{4{\pi }^{2}R}{T^2}\\ \hat{\phi }& 0=0\end{array}$

$T=\sqrt{\frac{4{\pi }^{2}R\tan \alpha }{g}}$     Note that T is independent of mass.