The weight of the meterstick is:
![W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N](https://tex.z-dn.net/?f=W%3Dmg%3D0.20%20kg%20%5Ccdot%209.81%20m%2Fs%5E2%20%3D%201.97%20N)
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
![d_1 = 0.50 m - 0.40 m=0.10 m](https://tex.z-dn.net/?f=d_1%20%3D%200.50%20m%20-%200.40%20m%3D0.10%20m)
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
![M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm](https://tex.z-dn.net/?f=M_w%20%3D%20W%20d_1%20%3D%20%281.97%20N%29%280.10%20m%29%3D0.20%20Nm)
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
![(mg) d_2 = 0.20 Nm](https://tex.z-dn.net/?f=%28mg%29%20d_2%20%3D%200.20%20Nm)
from which we find the value of d2:
![d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m](https://tex.z-dn.net/?f=d_2%20%3D%20%20%5Cfrac%7B0.20%20Nm%7D%7Bmg%7D%3D%20%5Cfrac%7B0.20%20Nm%7D%7B%280.5%20kg%29%289.81%20m%2Fs%5E2%29%7D%3D0.04%20m%20%20)
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
tension in rope = 25.0 N
Explanation:
- Two forces act on the suspended weight. The force coming down is the gravitational force and the upward force by the tension in the rope.
- Since the suspended weight is not accelerating so that the net force will be zero. Therefore the tension in the rope should be 25 N.
∑F = F - W = 0
so
F = W
so tension in rope = F = T = 25 N
Answer:
The combining of light nuclei is called nuclear fusion.
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
![Fn_{3} = \sqrt{405^{2}+701.5^{2} }](https://tex.z-dn.net/?f=Fn_%7B3%7D%20%3D%20%5Csqrt%7B405%5E%7B2%7D%2B701.5%5E%7B2%7D%20%20%7D)
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N