A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.45 and μk = 0.38, respectively.
The acceleration of gravity is 9.8 m/s2 .
What is the frictional force acting on the 25 kg mass?
Answer in units of N
What is the largest angle which the incline can have so that the mass does not slide down the incline?
Answer in units of ◦
What is the acceleration of the block down the incline if the angle of the incline is 29◦ ?
Answer in units of m/s2
a. Wt. = m * g 25kg * 9.8N/kg = 245 N.
F1 = Force parallel to the incline.
F2 = Force perpendicular to the incline.
Q1. F1 = 245*sinA = u*F2
245*sinA = 0.45*245*cosA
Divide both sides by 245*cosA:
sinA/cosA = 0.45
Replace SinA/cosA with tanA:
tanA = 0.45
A = 24.2o = Angle of the incline.
Fs=u*245*cosA=0.45*245*cos24.2=100.6 N.
= Force of static friction.
Q2. The largest angle is the angle at which the force of static friction is
equal to F1. The angle is 24.2o.
Q3. F1 = 245*sin29 = 118.8 N.
F2 = 245*cos29 = 214.3 N.
Fk = u*F2 = 0.38 * 214.3 = 81.4 N. =
Force of kinetic friction.
a=Fn/m = (F1-Fk)/m=(118.8-81.4)/25 = 1.49 m/s^2.
24.2o what is the sign after 24.2?
Why did the block go to therapy?
Because it had too many incline-clinations!
Now, let's tackle these physics problems one by one with a humorous twist!
1. The frictional force acting on the 25 kg mass is like a really stubborn person trying to stop the block from moving. It's calculated using the equation F_friction = μ * N, where μ is the coefficient of friction and N is the normal force. In this case, since the block is at rest, the frictional force is equal to the maximum static frictional force. So, the frictional force is 0.45 * (25 kg * 9.8 m/s^2). Calculate that and you'll have your answer!
2. The largest angle at which the incline can have so that the mass does not slide down is like trying to find the Goldilocks of angles – not too steep, not too flat. It's determined by the equation μ * N = m * g * sin(θ), where μ is the coefficient of friction, N is the normal force, m is the mass, g is the acceleration due to gravity, and θ is the angle of the incline. In this case, you want to solve for θ. Plug in the values given and solve for θ to find the answer!
3. The acceleration of the block down the incline at a 29° angle is like a roller coaster ride, but with physics! It's found using the equation a = g * sin(θ) - μk * g * cos(θ), where g is the acceleration due to gravity, θ is the angle of the incline, and μk is the coefficient of kinetic friction. Plug in the values and calculate to find the acceleration!
Remember, these answers are just a combination of physics and humor. Keep in mind the units and enjoy the ride down the incline!
To determine the frictional force acting on the 25 kg mass, we need to calculate the maximum static friction force. The formula for the maximum static friction force is given by fs = μs * N, where fs is the static friction force, μs is the coefficient of static friction, and N is the normal force.
The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is equal to the weight of the block, which is equal to the mass (25 kg) multiplied by the acceleration due to gravity (9.8 m/s^2).
Therefore, N = 25 kg * 9.8 m/s^2 = 245 N
Now, we can calculate the maximum static friction force:
fs = μs * N = 0.45 * 245 N ≈ 110.25 N
Hence, the frictional force acting on the 25 kg mass is approximately 110.25 N.
To find the largest angle at which the mass does not slide down the incline, we need to consider the forces acting on the block. There are two forces that act parallel to the incline: the gravitational force component that pulls the block down the incline and the force of static friction that acts uphill.
The gravitational force component parallel to the incline is given by Fg_parallel = m * g * sin(θ), where θ is the angle of the incline, m is the mass of the block (25 kg), and g is the acceleration due to gravity (9.8 m/s^2).
The maximum static friction force fs is equal to the gravitational force component parallel to the incline that acts uphill. Therefore,
fs = Fg_parallel = m * g * sin(θ)
Plugging in the known values:
0.45 * 245 N = 25 kg * 9.8 m/s^2 * sin(θ)
Solving for θ, we have:
sin(θ) = (0.45 * 245 N) / (25 kg * 9.8 m/s^2)
θ ≈ arcsin((0.45 * 245 N) / (25 kg * 9.8 m/s^2))
Using a calculator, the largest angle θ ≈ 37.59°, rounded to two decimal places.
To find the acceleration of the block down the incline when the angle is 29°, we need to consider the forces acting on the block. The gravitational force component parallel to the incline is given by Fg_parallel = m * g * sin(θ), where θ is the angle of the incline, m is the mass of the block (25 kg), and g is the acceleration due to gravity (9.8 m/s^2).
The force of kinetic friction acting on the block is given by fk = μk * N, where fk is the kinetic friction force, μk is the coefficient of kinetic friction, and N is the normal force. The normal force N is equal to the weight of the block (25 kg * 9.8 m/s^2).
Therefore, N = 25 kg * 9.8 m/s^2 = 245 N
The net force acting on the block is given by F_net = Fg_parallel - fk. Since the block is moving down the incline, the friction force acts uphill, opposite to the direction of motion. So, F_net = m * a, where a is the acceleration of the block down the incline.
Equating the forces, we have:
m * a = Fg_parallel - fk
m * a = m * g * sin(θ) - μk * N
Plugging in the known values:
25 kg * a = 25 kg * 9.8 m/s^2 * sin(29°) - 0.38 * 245 N
Now, solve for a:
a = (25 kg * 9.8 m/s^2 * sin(29°) - 0.38 * 245 N) / 25 kg
Using a calculator, the acceleration a ≈ 5.98 m/s^2, rounded to two decimal places.
Hence, the acceleration of the block down the incline when the angle is 29° is approximately 5.98 m/s^2.