A worker pushes horizontally on a 34 kg crate with a force of magnitude 110 n . the coefficient of static friction between the c
1 answer:
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Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is
Explanation:
From the question we are told that
The mass of the wheel is m = 6.9 kg
The radius is
The radius of gyration is
The angle is
The force which the massless bar is subjected to
Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that point A is the center of rotation.
Generally the moment of inertia about A is mathematically represented as
Here is the moment of inertia about G with respect to the radius of gyration which is mathematically represented as
=>
=>
=>
Generally the torque experienced by the wheel is mathematically represented as
=>
=>
Generally this torque is also mathematically represented as
=>
=>
Answer:
0.47 N
Explanation:
Here we have a ball in motion along a circular track.
For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.
This force is called centripetal force, and its magnitude is given by:
where
m is the mass of the object
is the angular velocity
r is the radius of the circle
For the ball in this problem we have:
m = 40 g = 0.04 kg is the mass of the ball
is the angular velocity
r = 30 cm = 0.30 m is the radius of the circle
Substituting, we find the force: