The question is incomplete. The complete question is :
To measure the effective coefficient of friction in a bone joint, a healthy joint (and its immediate surroundings) can be removed from a fresh cadaver. The joint is inverted, and a weight is used to apply a downward force F⃗ d on the head of the femur into the hip socket. Then, a horizontal force F⃗ h is applied and increased in magnitude until the femur head rotates clockwise in the socket. The joint is mounted in such a way that F⃗ h will cause clockwise rotation, not straight-line motion to the right. The friction force will point in a direction to oppose this rotation.
Draw vectors indicating the normal force n⃗ (magnitude and direction) and the frictional force f⃗ f (direction only) acting on the femur head at point A.
Assume that the weight of the femur is negligible compared to the applied downward force.
Draw the vectors starting at the black dot. The location, orientation and relative length of the vectors will be graded
Solution :
The normal force represented by N is equal to the downward force, 
which is equal in magnitude but it is opposite in direction.
Also the frictional force acts always to oppose the motion because the bone starts moving in a clockwise direction. The frictional force that will be applied to the right direction so that the movement or the rotation at A is opposed.
Answer:
F = -319.2 N
Explanation:
Given that,
The mass of a bicyclist, m = 70 kg
Mass of the bicycle = 9.8 kg
The speed of a bicycle, v = 16 m/s
We need to find the magnitude of the braking force of the bicycle come to rest in 4.0 m.
The braking force is given by :
So, the required force is 319.2 N.
Answer:
Conductors have magnetic fields; insulators do not have magnetic fields. Conductors do not have magnetic fields; insulators do have magnetic fields. ... In a conductor, electric current cannot flow freely; in an insulator, it can flow freely.
Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
The force on the layer will be equivalent to the weight of water on it. This is given by:
F = mg; m is the mass of water and g is the acceleration due to gravity.
