Question

a metal bar 70cm long and 4.00 kg in mass supported on two knief -edge placed 10 cm from each end. a 6.00 kg weight is suspended at 30 cm from one end . find the reaction at the knife-edge ?

Answer 1

The reactions at the knife-edges supporting the metal bar are $\boxed{56\,{\text{N}}\,{\text{\& }}\,{\text{44}}\,{\text{N}}}$ .

Further Explanation:

Since the metal bar is balanced at the knife edges. The net torque about the two knife edges will be zero.

Consider the reaction force at the first knife edge is ${F_1}$ and at the second knife edge is ${F_2}$ as shown in the figure below.

The weight of the metal ball is acting at the center of the bar i.e. $35{\text{ cm}}$ from the end of the bar. The weight of the metal bar is:

$\begin{aligned}W&= mg\\&=4\times10\\&= 40\,{\text{N}}\\\end{aligned}$

The reaction of the $6\,{\text{kg}}$ mass is:

$\begin{aligned}W'&=m'g\\&=6\times 10\\&= 60\,{\text{N}}\\\end{aligned}$

The net torque about the point $A$ can be expressed as:

$\begin{aligned}{F_2}\times 0.5 &=\left({60 \times 0.3} \right)+\left({40 \times 0.25} \right)\\{F_2}&=\frac{{18 + 10}}{2}\\&= 56\\\end{aligned}$

The net torque about the point   can be expressed as:

$\begin{aligned}{F_1}\times 0.5 &= \left({60 \times 0.2}\right) + \left({40 \times 0.25} \right)\\{F_1}&=\frac{{12 + 10}}{2}\\&=44\\\end{aligned}$

Thus, the reactions at the knife-edges supporting the metal bar are $\boxed{56\,{\text{N}}\,{\text{\& }}\,{\text{44}}\,{\text{N}}}$.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Torque and Moment

Keywords:  Metal bar, 70cm long, supported, two knife-edge, 10cm from each end, weight is suspended, reaction at, net torque, center of metal bar.

Answer 2

First, you make a diagram of all the forces acting on the system. This is shown in the figure. We have to determine F1 and F4. Let's do a momentum balance. Momentum is conserved so the summation of all momentum is equal to zero. Momentum is force*distance.

To determine F1: (reference is F4, so F4=0)

∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)

To determine F4: (reference is F1, so F1=0)

∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge)