Question

The slotted arm revolves in the horizontal plane about the fixed vertical axis through point O. The 2.6-lb slider C is drawn toward O at the constant rate of 2.1 in./sec by pulling the cord S. At the instant for which r = 6.1 in., the arm has a counterclockwise angular velocity ω = 4.2 rad/sec and is slowing down at the rate of 3.6 rad/sec 2. For this instant, determine the tension T in the cord and the force N exerted on the slider by the sides of the smooth radial slot. The force N is positive if side A contacts the slider, negative if side B contacts the slider.

Answer 1

Answer:

Explanation:

Given

Weight of slider=2.6 lb

radius(r)=6.1 in.

$\omega (\dot{\theta })=4.2 rad/s$

$\ddot{\theta }=3.6 rad/s^2$

$\dot{r}=2.1$

From FBD we can say that

$\sum F_r=ma_r=m\left ( \ddot{r}-r\dot{\theta }^2\right )$

$-T=\frac{2.6}{32.2}\left ( 0-\frac{6.1}{12}\times 4.2^2\right )$

T=0.726 lb

$\sum F_{\theta }=ma_\theta=m\left ( r\ddot{\theta }+2\dot{r}\dot{\theta }\right )$

$N=\frac{2.61}{32.2}\left [ \frac{6.1}{12}\times \left ( -2\right )+2\left ( -\frac{2.1}{12} \right )4.2\right ]$

N=-0.201 lb(contact on B side)