A 200-kg object and a 500-kg object are separated by 4.00 m. (a) find the net gravitational force exerted by these objects on a
50.0-kg object placed midway between them. (b) at what position (other than an infinitely remote one) can the 50.0-kg object be placed so as to experience a net force of zero from the other two objects?
1 answer:
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Given :
Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south.
To Find :
The magnitude and direction of
a) A + B .
b) A - B.
Solution :
Let , direction in north is given by +j and east is given by +i .
So , and
Now , A + B is given by :
Direction of A+B is 45° north of west .
Also , for A-B :
Direction of A-B is 45° south of west .
( When two vector of same magnitude which are perpendicular to each other are added or subtracted the resultant is always 45° from each of them)
Hence , this is the required solution .
Answer: Add an incline or grade to the road track.
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).