Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for DAC:
⇒
Now for the BAC:
⇒
Now, differentiating w.r.t x:
For maximum angle, = 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 =
After solving the above eqn, we get
x =
The observer should stand at a distance equal to x =
Answer:
the equivalent mass :
the equation of the motion of the block of mass in terms of its displacement is =
Explanation:
Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder
The radius of the cylinder be = R
The distance between the center of the pulley to center of the block to be = x
Also, the angles of inclinations of the cylinder and the block with respect to the ground to be and
respectively.
The velocity of the block to be = v
The equivalent mass of the system =
In the terms of the equivalent mass, the kinetic energy of the system can be written as:
--------------- equation (1)
The angular velocity of the cylinder = : &
The inertia of the cylinder about its center to be = I
The angular velocity of the cylinder can be written as:
The kinetic energy of the system in terms of individual mass can be written as:
By replacing with
; we have:
------------------ equation (2)
Equating both equation (1) and (2); we have:
Therefore, the equivalent mass : which is read as;
The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.
The expression for the force acting on equivalent mass due to the block is as follows:
Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:
Equating the above both equations; we have the equation of motion of the equivalent system to be
which can be written as follows from the previous derivations
Finally; the equation of the motion of the block of mass in terms of its displacement is =
Answer: 12 N to the right
Explanation:
If we calculate the net force acting on the box, we will have:
<u>In y-component:</u>
(1)
Where is the Normal force, directed upwards and
is the weight of the box (gravity force), directed downwards.
(2)
(3) Hence the net force in the vertical component is zero
<u>In x-component:</u>
(4)
Where and
(5)
(6) This is the net force in the horizontal component
Therefore, the total net force acting on the box is 12 N directed to the right
Answer:
mu = 0.56
Explanation:
The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:
v: final speed = 0m/s (the car stops)
v_o: initial speed in the interval of interest = 60km/h
= 60(1000m)/(3600s) = 16.66m/s
x: distance = 25m
BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:
with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:
Furthermore, you use the relation between the friction force and the friction coefficient:
hence, the friction coefficient is 0.56