Respuesta :
Answer:
(a) Tension on both side of wire
[tex]T_1=46.706 N[/tex]
[tex]T_2=51.710 N[/tex]
(b) acceleration of the Box
[tex]a = 3.336 \frac{m}{sec^2}[/tex]
(c) The horizontal and vertical components
Horizontal component [tex]T_1=46.706 N[/tex]
Vertical Component =130.19 N
Explanation:
Refer attached figure for details.
[tex]T_1\ \&\ T_2\ are\ tensions\ in\ the\ string\ and\ a\ is\ the\ acceleration\ of\ the\ masses.[/tex]
Applying Newton's 2 law of motion for 14 kg block in horizontal direction
[tex]T_1 = 14\ a[/tex]-----------(i)
Similarly, applying Newton's 2 law of motion for 8 kg block in vertical direction
[tex]8 g - T_2 =8 a[/tex]-----(ii)
Consider the case of pulley,
[tex]\tau_e_x_t= I\alpha--------(iii)\\\\Where,\\\tau_e_x_t =Torque\ acting\ on\ the\ pulley\\I=moment\ of\ inertia\ of\ pulley\\\alpha= angular\ acceleration[/tex]
where,
[tex]I= \frac{MR^2}{2} (for\ pulley\ disk)[/tex]
[tex]I=\frac{3\cdot0.5^2}{2} =0.375\ kgm^2[/tex] (since mass of the pulley = 3 kg & Radius = 0.5 m)
&[tex]\tau_e_x_t= Net\ force \cdot Distance\ from\ application\ point[/tex]
Hence [tex]\tau_e_x_t = (T_2-T_1) \cdot \frac{1}{2} = 0.375\cdot\alpha[/tex]
[tex]T_2-T_1=0.75\cdot\alpha[/tex]--------(iv)
Relation between linear acceleration (a) and angular acceleration (α) is as follows,
[tex]a = R\alpha=0.5\cdot\alpha \ (R\ is \ radius\ of \ pulley)[/tex]
[tex]\alpha=2a[/tex]--------------------(v)
Putting the value of (v) in to (iv)
[tex]T_2 -T_1= 1.5 a[/tex]---------(vi)
adding equation (i),(ii) & (vi) gives
8g =22 a + 1.5 a
[tex]a = 3.336 \frac{m}{sec^2}[/tex]
now putting the value of a in equation (i) & (ii) we get
[tex]T_1=46.706 N[/tex]
[tex]T_2=46.706 +1.5 \cdot 3.336[/tex] = 51.710 N
(a) Hence Tension on both side of wire
[tex]T_1=46.706 N[/tex]
[tex]T_2=51.710 N[/tex]
(b) acceleration of the Box
[tex]a = 3.336 \frac{m}{sec^2}[/tex]
(c) The horizontal and vertical components
Horizontal component [tex]T_1=46.706 N[/tex]
Vertical Component = [tex]T_2+8\cdot g[/tex] =51.710 + 8 x 9.81 =130.19 N
Answer:
Tension, T1 = 46.2 N and T2 = 52 N, where as acceleration = 3.3 ms^-2.
Forces on the pulley are 46.2 N , 81.4 N horizontal and vertical
respectively.
Explanation:
Given:
Mass of the box on rest,[tex]m_1[/tex] = 14 kg
Mass of the attached box,[tex]m_2[/tex] = 8 kg
Mass of the pulley, [tex]m_3[/tex] = 3 kg
Diameter of the pulley, [tex]d[/tex] = 1 m
Radius of the pulley, [tex]r[/tex] = 0.5 m
Here we will be using the concept of net force ([tex]F_n_e_t[/tex]),net torque ([tex]\tau_n_e_t[/tex]) and acceleration of the pulley .
A FBD is attached with.
Lets find the tension on the wire using Fnet.
⇒ [tex]T_1=m_1(a)[/tex] ...for m1
⇒ [tex]m_2g-T_2=m_2(a)[/tex] can be written as [tex]T_2=m_2g-m_2(a)[/tex] ...for m2
Considering clockwise torque as negative and anticlockwise torque as positive.
Moment of inertia (I) of the disk/pulley = [tex]\frac{m_3r^2}{2}[/tex] and [tex]\alpha=\frac{a}{r}[/tex] .
Now using net torque on the pulley we can say that.
⇒ [tex](T_2-T_1)r=I\alpha[/tex]
⇒ [tex](T_2-T_1)r=\frac{m_3r^2}{2}\times \frac{a}{r}[/tex]
⇒ [tex](T_2-T_1)=\frac{m_3a}{2}[/tex]
⇒ Plugging T1 and T2 .
⇒ [tex]m_2g-m_2a-m_1a=\frac{m_3a}{2}[/tex]
⇒ Isolating a from the rest.
⇒ [tex]m_2g=\frac{m_3a}{2}+m_2a+m_1a[/tex]
⇒ [tex]m_2g=a\ [\frac{m_3}{2}+m_2+m_1][/tex]
⇒ [tex]\frac{m_2g}{\frac{m_3}{2} +m_2+m_1} =a[/tex]
⇒ Plugging the numeric value
⇒ [tex]\frac{(8\times 9.8)}{(\frac{3}{2} +8+14)} =a[/tex]
⇒ [tex]3.3 =a[/tex]
⇒ Acceleration = 3.3 [tex]ms^-^2[/tex]
So,
(a).
Tension in the wire
⇒ [tex]T_1=m_1(a)=14\times 3.3 =46.2\ N[/tex]
⇒ [tex]T_2=m_2g-m_2(a)=8(9.8-3.3)=52\ N[/tex]
(b).
The acceleration of the box is 3.3 ms^-2.
(c).
Forces on the pulley.
Horizontal force, [tex]P_H[/tex] = [tex]T_1[/tex] = [tex]46.2\ N[/tex]
Vertical force,[tex]P_V[/tex] = [tex]T_2+m_3g[/tex] = [tex]52+3(9.8)[/tex] = [tex]81.4\ N[/tex]
The values are as follows:
Tension as T1 = 46.2 N and T2 = 52 N ,where as acceleration =3.3 ms^-2.
Forces on the pulley are 46.2 N , 81.4 N horizontal and vertical
respectively.
