Question

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what distance from the left end of the rod should a 0.6kg mass be hung to balance the rod

Answer 1

Answer:

45cm

Explanation:

given

length of rod=50cm

mass of rod= 0.2kg

kindly find attached the free body diagram

we know that clockwise moment = anticlockwise moment

taking moment about the fulcrum we have

0.2*15cm=0.6*x

3=0.6x

x=3/0.6

x=5cm

since the 0.6kg mass is place 5cm from the fulcrum which is 40cm from the end of the rod. the total distance of the 0.6kg mass from the left end is

5+40=45cm

τcw=r∗F=r(10ms2)(0.6kg)=0.3N∗m.

Solving for r gives r = 0.05m to the right of the pivot, so 40 + 5 cm from the left end of the rod.

Answer 2

The distance from the left end of the rod at which mass B must be hung to balance the rod is 45 cm.

Given the following data:

• Mass of rod = 0.2 kg

• Length of rod = 50 cm to m = 0.5 meter

• Distance = 40 cm to m = 0.4 meter

• Mass B = 0.6 kg

To determine the distance from the left end of the rod at which mass B must be hung to balance the rod:

For a rod to be balanced, the clockwise and counterclockwise torques about the pivot point must have the same magnitude i.e they must be equal in magnitude.

Hence, using the pivot point as a reference, the center of the rod is 15 cm from the reference.

$0.2 \times 15 = 0.6 \times b\\\\3 =0.6b\\\\b=\frac{3}{0.6}$

b = 5 cm

Next, we would calculate the counterclockwise torque by using the formula:

$T_{ccw} = Fr\\\\T_{ccw} = mgr\\\\T_{ccw} = 0.2 \times 10 \times 0.15 \\\\T_{ccw} = 0.3 \;Nm$

For the clockwise torque:

$T_{cw} = Fr\\\\T_{cw} = mgr\\\\T_{cw} = 0.6 \times 10 \times r \\\\T_{cw} = 6r\;Nm$

$T_{ccw}=T_{cw}\\\\0.3 =6r\\\\r=\frac{0.3}{6}$

r = 0.05 meter

In centimeter:

r = 5 cm

From the left end of the rod and to its right end:

$Distance = 5 + 40$

Distance = 45 cm

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