Answer:
The frictional torque is 
Explanation:
From the question we are told that
The mass attached to one end the string is 
The mass attached to the other end of the string is 
The radius of the disk is 
At equilibrium the tension on the string due to the first mass is mathematically represented as
substituting values
At equilibrium the tension on the string due to the mass is mathematically represented as
The frictional torque that must be exerted is mathematically represented as
substituting values
Hi hi I hope you know that you have a great time jaha
Rise over run at 1 second
It’s the same slope from 0 to 2 seconds
10/2=5mps
As a note all time points between 0and 2 will have this instantaneous velocity
Instantaneous velocity at time 2 is 0
6. Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.
F = ma
F = (15.3)(-9.8)
F = -150N
Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object. So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.
7. Same idea as question 2.
First determine the weight of the object:
F = ma
F = (30)(-9.8)
F = -294N
The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.
-294 + 150N + x = 0
x = 144N
So the person is exerting 144 N.
10. First find the force of block B to the right due to its acceleration:
F = ma
F = (24)(0.5)
F = 12N
So block B is moving 12N to the right relative to block A due to block A's movement to the left. However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A. The force that is causing block B to experience the lower relative force to the right is because of the friction. To find the friction:
The sum of the forces in the leftward and rightward direction for block B must equal 12N.
75 - x = 12
x = 63N
So the force of friction of block A on block B is 63N to the left.
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
