A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache
d to a vertical wall by a string 6 meter long and 5 meter away from the wall. Find the magnitude of the horizontal force F,APPLIED TO THE LOWER block that shall be necessary so that slipping of 100 kg block occurs. (take coefficient of friction for both contacts =0.25 )
1 answer:
Answer:
Explanation:
<u>Friction Force</u>
When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.
There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.
Please find the free body diagrams in the figure provided below.
The equilibrium condition for the mass 1 is
The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa
The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is
The friction forces are computed by
Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.
Replacing in [1]
Simplifying
Plugging in the values
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A) t = 0 s, 4 s, 8 s
B) t = 2 s, 6 s
C) t = 1 s, 3 s, 5 s, 7 s
Explanation:
A)
The figure is missing: find it in attachment.
A particle is said to be in Simple Harmonic Motion (SHM) when it is acted upon a restoring force proportional to its displacement, and therefore, the acceleration of the particle is directly proportional to its displacement (but in the opposite direction):
Also, the displacement of a particle in SHM can be described by a sinusoidal function, as shown in the figure for the particle in this problem.
For the particle in this problem, from the figure we can see that the displacement is described by a function in the form
where
A is the amplitude
is the angular frequency, where T is the period. From the graph, we see that the particle completes 1 oscillation in 4 seconds, so
T = 4 s
So the angular frequency is
So
The velocity of a particle in SHM can be found as the derivative of the displacement, so here we find:
So, the particle is moving to the right at maximum speed when
(because this is the maximum positive value for the cosine part). Solving for t,
We also have another time in which this occurs, when
And also when
B)
In this case, we want to find the time t at which the particle is moving to the left at maximum speed.
This occurs when the cosine part has the maximum negative value, so when
Which means that this occurs when:
Also when
So, the particle has maximum speed moving to the left at 2 s and 6 s.
C)
The particle is instantaneously at rest when the speed is zero, this means when the cosine part is equal to zero:
This occurs when the argument of the cosine is:
Also when
And when
And finally when
So, the particle is at rest at t = 1 s, 3 s, 5 s, 7 s.
Work done by a given force is given by
here on sled two forces will do work
1. Applied force by Max
2. Frictional force due to ground
Now by force diagram of sled we can see the angle of force and displacement
work done by Max =
Now similarly work done by frictional force
Now total work done on sled
Answer:
1)
2)
Explanation:
<u>Projectile Motion</u>
When an object is launched near the Earth's surface forming an angle with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.
The heigh of an object can be computed as
Where is the initial height above the ground level,
is the vertical component of the initial velocity and t is the time
The y-component of the speed is
1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of
The object will reach the maximum height when . It allows us to compute the time to reach that point
Solving for
Thus, the maximum heigh is
We know this value is 8 meters
Solving for
Replacing the known values
2) We know at t=1.505 sec the ball is above Julie's head, we can compute