A boat can travel 2.30 m/s in still water. If the boat points its prow directly across a stream whose current is 1.80 m/s , what
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Answer:
a. Fnet =37.67N
b. The direction = 133.4 from the x axis counter clockwise.
c. Option 2
Explanation:
Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.
Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..
Given that F1 is 71N at 20°, then it is in the first quadrant.
a. Fnet= F1+F2+F3
Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j
Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j
Resolving the vectors into x and y components.
Fnet= -2.04i+37.62j
Magnitude of the vector
Fnet= √((-2.04)^2+(37.62)^2)
Fnet= 37.67N
Fnet is approximately 38N.
b. Direction of the Fnet.
Angle=arctan(y/x)
Angle=arctan(-37.61/2.04)
Angle= -43.37°
The angle is in the negative x axis and positive y axis.
Then the direction becomes 180-43.37
Therefore, the direction of the net force is 133.37°.
c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.
Fnet=ma
Fnet= mv/t
So the velocity is in the direction of the Fnet.
Answer:
a) 20 rad/s
b) 6 m/s
Explanation:
b) Force acting on the wheel is 200 N
mass of the wheel is 100 kg
From Newton's second law of motion, F = m × a
Where F is the net force acting on the body
m is mass of the body
a is the acceleration of the body
By substituting the values we get, a = 2 m/s²
As acceleration is constant, we can use the below formula for calculating the final velocity of the object
v = u + a × t
Where v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken
u = 0 (∵ it starts from rest)
By substituting the values we get
v = 0 + 2 × 3 = 6 m/s
∴ Speed of center of mass after 3 seconds = 6 m/s
a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel
∴ Radius of the wheel = 300 mm
Torque acting on the wheel about axis of rotation = 300 mm × 200 N =
60 N·m
Torque = (Moment of inertia) × (angular acceleration)
Assuming that the mass of spokes of the wheel to be negligible,
Moment of inertia of the wheel about axis of rotation = 100 × 300² × = 9 kg·m²
Then,
60 = 9 × (angular acceleration)
∴ angular acceleration ≈ 6·67 rad/s²
As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed
=
+ α × t
Where
is the final angular velocity
is the initial angular velocity
α is the angular acceleration
t is the time taken
is 0 (∵ initially it starts from rest)
By substituting the values we get
= 6·67 × 3 = 20 rad/s
∴ Angular velocity of the wheel after three seconds = 20 rad/s