The strength of the gravitational field is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
And now we can use the previous equation to calculate the field strength at that altitude:
And we can see this value is a bit less than the gravitational strength at the surface, which is
.
Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N
If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Given the data in the question;
- Length of the massless beam;
- Distance of support from the left end;
- First mass;
- Distance of beam from the left end( m₁ is attached to );
- Second mass;
- Distance of beam from the right of the support( m₂ is attached to );
Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.
Hence, 
we divide both sides by 
Next, we make 
, the subject of the formula
![x_1 = x - [ \frac{m_2x_2}{m_1} ]](https://tex.z-dn.net/?f=x_1%20%3D%20x%20-%20%5B%20%5Cfrac%7Bm_2x_2%7D%7Bm_1%7D%20%5D)
We substitute in our given values
![x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]](https://tex.z-dn.net/?f=x_1%20%3D%203.00m%20-%20%5B%20%5Cfrac%7B61.7kg%5C%20%2A%20%5C%200.273m%7D%7B31.3kg%7D%20%5D)
Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m
Learn more; brainly.com/question/3882839
Answer:
Explanation:
In order to solve this problem we need to make a free body diagram of the book and the forces that interact on it. In the picture below you can see the free body diagram with these forces.
The person holding the book is compressing it with his hands, thus exerting a couple of forces of equal magnitude and opposite direction with value F.
Now the key to solving this problem is to analyze the equilibrium condition (Newton's third law) on the x & y axes.
To find the weight of the book we simply multiply the mass of the book by gravity.
W = m*g
W = 1.3[kg] * 9.81[m/s^2]
W = 12.75 [N]
