Question

e 18-19 - marder - (Alcala 302K_1) 0 1 2 3 4 5 6 7 9 kg What is the weight of the measuring stick if it is balanced by a support force at the 1 m mark? The acceleration of gravity is 9.81 m/s. Answer in units of N. 013 10.0 points An Atwood machine is constructed using two v-wheels (with the masses concentrated at the rims). The left wheel has a mass of 2.2 kg and radius 23.41 cm. The right wheel has a mass of 2.6 kg and radins 28.I cm. The hanging mass on the left is 1.97 kg iind on the right 1.31 kg vity rm

Answer 1

The weight of the ball hanged at left end of measuring stick is = $\dpi{100} 9g$ = 9 x 9.81 = 88.29 N

Let the mass of the stick is m kg. So weight of the stick is mg N .

We know that , the weight of the any object is centred at the center of mass (theoretically) of the object.

As the total length of the stick is 7 m , so the center of mass (COM) of the stick will be at 3.5 m.

Therefore, the 9 kg ball is 1 m away from the support force and the ' m ' kg stick is 2.5 m away from the support force.

As the system is in rotational equilibrium then we can write ,

Torque of stick about the support = Torque of ball about the support

or, 2.5 x m x 9.81 = 1 x 88.29

or, m = 9/2.5 = 3.6 kg

So, weight of the stick is = 3.6 x 9.81 = 35.316 N.

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