A horse is harnessed to a sled having a mass of 236 kg, includ- ing supplies. The horse must exert a force exceeding 1240 N at a
n angle of 35.0° in order to get the sled moving. Treat the sled as a point particle.
(a) Calculate the normal force on the sled when the magnitude of the applied force is 1240 N.
(b) Find the coefficient of static friction between the sled and the ground beneath it.
(c) Find the static friction force when the horse is exerting a force of 6.20 x 10^2 N on the sled at the same angle.
(a) Calculate the normal force on the sled when the magnitude of the applied force is 1240 N.
(b) Find the coefficient of static friction between the sled and the ground beneath it.
(c) Find the static friction force when the horse is exerting a force of 6.20 x 10^2 N on the sled at the same angle.
2 answers:
Answer:
a) The normal force on the sled is 1.61 × 10³ N when the applied force is 1240 N at 35°.
b) The coefficient of static friction is 0.631
c) The friction force is 1.24 × 10³ N when the horse is exerting a force of 6.20 × 10² N at 35°
Explanation:
Hi there!
a) Since the sled is not moving in the vertical direction, the sum of the forces applied in this direction is zero.
Considering the upward direction as positive, the sum of vertical forces will be (see figure):
∑F = Fy + N - W = 0
Where:
Fy = vertical component of the force applied by the horse.
N = normal force on the sled.
W = weight of the sled.
Solving the equation for N:
N = W - Fy
The weight is the force that gravity exerts on the sled:
W = m · g
Where:
m = mass of the sled.
g = acceleration due to gravity (9.81 m/s²).
Then:
W = 236 kg · 9.81 m/s² = 2.32 × 10³ N
The y-component of the force applied by the horse is calculated using trigonometry:
sin θ = opposite / hypotenuse
sin 35.0° = Fy / F
sin 35.0° = Fy / 1240 N
1240 N · sin 35.0° = Fy
Fy = 711 N
Then, the normal force will be:
N = W - Fy
N = 2.32 × 10³ N - 711 N = 1.61 × 10³ N
The normal force on the sled is 1.61 × 10³ N when the applied force is 1240 N at 35°.
b) For the sled to start moving, the exerted force by the horse has to be greater than the static friction force (Fr). We know from the problem that the horse must exert a force greater than 1240 N at 35.0° to start moving the sled. Then, the magnitude of the friction force is equal to the magnitude of the horizontal component of the applied force, Fx (see figure).
Fx = 1240 N · cos 35°
Fx = 1016 N
Then the friction force is Fr = 1016 N.
The static friction force is calculated as follows:
Fr = μ · N
Where:
μ = coefficient of static friction
N = normal force
Then, solving for μ:
Fr / N = μ
1016 N / 1.61 × 10³ N = μ
μ = 0.631
The coefficient of static friction is 0.631
c) Now, the force exerted by the horse is 6.20 × 10² N.
So, let´s find again the vertical component of this force to calculate the normal force:
Fy = F · sin θ
Fy = 6.20 × 10² N · sin 35.0°
Fy = 356 N
The normal force will be:
N = W - Fy
N = 2.32 × 10³ N - 356 N
N = 1.96 × 10³ N
The friction froce will be:
Fr = μ · N
Fr = 0.63 · 1.96 × 10³ N = 1.24 × 10³ N
The static friction force is 1.24 × 10³ N when the horse is exerting a force of 6.20 × 10² N at 35°
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Answer:
1817.448 m
Explanation:
Given:
Initial acceleration = 2.24 m/s²
Time for acceleration = 10 s
From Newton's equation of motion
v = u + at
here v is the final velocity after the acceleration
v = 0 + 2.24 × 10 = 22.4 m/s
thus,
the displacement during acceleration
From Newton's equation of motion
where,
s is the distance
u is the initial speed = 0 m/s as starting from rest
a is the acceleration
t is the time
on substituting the respective values, we get
or
s₁ = 112 m
for case 2
time = 1.25 minute = 1.25 × 60 = 75 seconds
displacement, s₂ = v × t = 22.4 × 75 = 1680 m
for the case 3
acceleration = - 9.86 m/s²
now,
the final velocity for the second case is the initial velocity for this case
thus,
or
s₃ = 50.848 - 25.40 = 25.448 m
hence,
the total displacement = s₁ + s₂ + s₃ = 112 m + 1680 m + 25.448 m
= 1817.448 m