Explanation:
The rod is uniform, so the center of gravity is at the center, or 0.75 m from the end. The wedge is 0.5 m from the end, so the center is 0.25 m from the wedge.
Sum the torques about the wedge (it may help to draw a diagram first). Take counterclockwise to be positive.
∑τ = Iα
W (0.25 m) − (100 N) (0.50 m) = 0
W = 200 N
Sum the forces in the y direction.
∑F = ma
F − 100 N − 200 N = 0
F = 300 N
Answer:

Explanation:
Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass
at a distance r will be:

where
is the gravitational constant.
This force is the centripetal force the satellite experiments, so we can write:

Putting all together:

which means:
![r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGM%7D%7B4%5Cpi%5E2%7DT%5E2%7D)
Which for our values is:
![r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%286.67%5Ctimes10%5E%7B-11%7DNm%5E2%2Fkg%5E2%29%286.39%5Ctimes10%5E%7B23%7D%20kg%29%7D%7B4%5Cpi%5E2%7D%281.026%5Ctimes24%5Ctimes60%5Ctimes60s%29%5E2%7D%3D20395282m%3D20395.3km)
Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km
, which leaves us with:

So Force=Mass*Acceleration
You have a mass of .175 kg or 175 grams
Force= (175 g)(50)= 8750 N of force
Hookes law states F=kX where F is the force applied, k is the spring constant, and X is the extension of the spring from its resting point.
Substituting the values in, we get:
F=100*(0.65-0.5)=100*0.15=15N