Question

A 1kg rod of length 1m is pivoted at its centre and two masses of 5kg and 2kg and are hung from the ends as shown in figure. Find the tension in the supports to the blocks of mass 2kg and 5kg $(g=9.8\frac{m}{s^2})$

• A. 20.6N, 29N
• B. 20.6N, 20.6N
• C. 27.6N, 20.6N
• D. 27.6N, 29N

Answer 1

The correct option is D

27.6N, 29N

In this problem the rod has a mass 1 kg

$a)t_{net}=1_{net}\times \alpha \Rightarrow 5\times 10\times 10.5-2\times 10\times 0.5\Rightarrow (5\times {\left(\frac{1}{2}\right)}^2+2\times {\left(\frac{1}{2}\right)}^2+\frac{1}{12})\times \alpha \Rightarrow 15=(1.75+0.084)\alpha \Rightarrow \alpha =\frac{1500}{(175+8.4)}=\frac{1500}{183.4}=8.1\frac{rad}{s^2}(g=10)=8.1\frac{rad}{s^2}(ifg=9.8)$

$a)T_1-m_{1}g=m_{1}a\Rightarrow T_1=m_{1}a+m_{1}g=2(a+g)=2(\alpha r+g)=2(8\times 0.5+9.8)$

= 27.6 N on the first body.

In the second body
$\Rightarrow m_{2}g-T_2=m_{2}a\Rightarrow T_2=m_{2}g-m_{2}a\Rightarrow T_2=5(g-a)=5(9.8-8\times 0.5)=29N.$