You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If
62-kg Dan sits on the left end of the board and 50-kg Tahreen on the right end of the board, where should 54-kg Komila sit to keep the board stable? Ignore the mass of the board and treat each student as point-like objects. The positive x-direction is to the right, and the origin is at the center of the chair.
1 answer:
Answer:
Komila should sit 0.33m from the middle of the board towards tahreen.
Explanation:
We are told to treat each student as point-like objects. So i have attached a rigid body diagram to depict this.
From the diagram,
F_d is force exerted by dan
F_t is force exerted by tahreen
F_k is force exerted by komila
F_b is force of board at the mid point.
x1 is distance of dan from the centre of the chair
x2 is distance of komila from the centre of the chair
x3 is distance of tahreen from komila
We are given;
Mass of Dan;m_d = 62 kg
Mass of tahreen;m_t = 50 kg
Mass of komila;m_k = 54 kg
Now, taking moments about the centre of the chair, we have;
(F_d*x1) - (F_k*x2) - (F_t(x2 + x3)) = 0
Now,F_d = m_d*g ; F_t = m_t*g ; F_k = m_k*g
We are told that the board is 3m long. So, if we assume that the fulcrum position of the chair coincides with the midpoint of boards length, we'll have;
x1 = (x2 + x3) = 1.5
Thus, we now have;
(F_d*1.5) - (F_k*x2) - (F_t*1.5) = 0
F_d = m_d*g = 62 * 9.8 = 607.6 N
F_t = m_t * g = 50 x 9.8 = 490 N
F_k = m_k * g = 54 x 9.8 = 529.2 N
So plugging in these values, we have;
(607.6 * 1.5) - (529.2 * x2) - (490 * 1.5) = 0
911.4 - 735 = 529.2 x2
529.2 x2 = 176.4
x2 = 176.4/529.2
x2 = 0.33m
Komila should sit 0.24m from the middle of the board towards tahreen
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Answer:
a) The maximum height reached by the projectile is 558 m.
b) The projectile was 21.3 s in the air.
Explanation:
The position and velocity of the projectile at any time "t" is given by the following vectors:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v = (v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time "t"
x0 = initial horizontal position
v0 = initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).
v = velocity vector at time t
a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:
vy = v0 · sin α + g · t
0 = 175 m/s · sin 36.7° - 9.80 m/s² · t
- 175 m/s · sin 36.7° / - 9.80 m/s² = t
t = 10.7 s
Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).
Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²
y = 558 m
The maximum height reached by the projectile is 558 m
b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.
We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²
0 = t (175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t)
0 = 175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t
- 175 m/s · sin 36.7 / -(1/2 · 9.8 m/s²) = t
t = 21.3 s
The projectile was 21.3 s in the air.
Answer:
the cannonball’s velocity parallel to the ground is 86.6m/S
Explanation:
Hello! To solve this problem remember that in a parabolic movement the horizontal component X of the velocity of the cannonball is constant while the vertical one varies with constant acceleration.
For this case we must draw the velocity triangle and find the component in X(see atached image).
V= Initial velocity=100M/S
V= Initial velocity=100M/S
Vx=cannonball’s velocity parallel to the ground
Solving for Vx
Vx=Vcos30
Vx=(100m/S)(cos30)=86.6m/s
the cannonball’s velocity parallel to the ground is 86.6m/S
