A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.67 and μk = 0.57, respectively.
The acceleration of gravity is 9.8 m/s2 .
-What is the frictional force acting on the 30 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 ◦
please show work and the answers need to be rounded to the third decimal place.
normal force = m g cos A
friction force max = μs m g cos A
component of weight down slope = m g sin A
slips when
m g sin A = μs m g cos A
or
tan A = μs
Well, well, well! Looks like you need my help with some physics problems. Don't worry, I've got just the right amount of humor to tackle this!
First, let's find the frictional force acting on the 30 kg mass. The frictional force can be calculated by multiplying the coefficient of static friction (μs) by the normal force (N). Here's the equation:
Frictional force = μs * N
Since the block is at rest, the normal force acting on it is equal to its weight. So, the normal force can be calculated by multiplying the mass (30 kg) by the acceleration due to gravity (9.8 m/s^2).
Normal force (N) = mass * gravity
Now that we know the normal force, we can find the frictional force:
Frictional force = 0.67 * (30 kg * 9.8 m/s^2)
Calculating that will give us the frictional force in newtons.
Now, let's move on to the second part of the question, finding the largest angle at which the mass does not slide down the incline. In order for the mass to not slide down, we need to consider the maximum static friction force, which can be calculated using the equation:
Maximum static friction force (Fs) = μs * N
Since the mass is at the brink of sliding down, the force acting downwards along the incline is equal to the maximum static friction force (Fs). This force can be calculated using the following equation:
Fs = mass * gravity * sin(θ)
Where θ is the angle of the incline. Rearranging the equation, we can find the largest angle:
θ = arcsin(Fs / (mass * gravity))
Now you've got all the tools to solve these physics problems while enjoying a good laugh!
To find the frictional force acting on the 30 kg mass, we need to consider the components of forces acting on the block on the incline.
1. Determine the normal force (N): The weight of the block is given by W = mg, where m is the mass (30 kg) and g is the acceleration due to gravity (9.8 m/s^2). Therefore, W = 30 kg × 9.8 m/s^2 = 294 N. The normal force N acting perpendicular to the incline is equal to the weight of the block since it is on a horizontal surface. Therefore, N = 294 N.
2. Determine the maximum static frictional force (Fs,max): The maximum static frictional force is given by Fs,max = μsN, where μs is the coefficient of static friction. In this case, μs = 0.67. Therefore, Fs,max = 0.67 × 294 N = 196.98 N.
3. Determine the kinetic frictional force (Fk): The kinetic frictional force is given by Fk = μkN, where μk is the coefficient of kinetic friction. In this case, μk = 0.57. Therefore, Fk = 0.57 × 294 N = 167.58 N.
Since the block is at rest, the frictional force acting on it must be equal to the maximum static frictional force. Therefore, the frictional force acting on the 30 kg mass is 196.98 N (rounded to the third decimal place).
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 along the incline.
1. Determine the force of gravity along the incline (Fg_parallel): The force of gravity acting on the block has two components: one perpendicular to the incline (W) and one parallel to the incline (Fg_parallel). The force of gravity along the incline can be calculated as Fg_parallel = mg × sin(θ), where θ is the angle of the incline. Plugging in the values, Fg_parallel = 30 kg × 9.8 m/s^2 × sin(θ).
2. Determine the maximum static frictional force (Fs,max) along the incline: The maximum static frictional force along the incline is given by Fs,max = μsN, where μs is the coefficient of static friction and N is the normal force (equal to the weight of the block). In this case, N = 30 kg × 9.8 m/s^2. Therefore, Fs,max = 0.67 × 30 kg × 9.8 m/s^2.
For the block not to slide down the incline, the force of gravity along the incline must be balanced by the maximum static frictional force along the incline. Therefore, we can set Fg_parallel equal to Fs,max and solve for the angle θ.
30 kg × 9.8 m/s^2 × sin(θ) = 0.67 × 30 kg × 9.8 m/s^2
Simplifying the equation:
sin(θ) = 0.67
Taking the inverse sine (sin^(-1)) of both sides:
θ = sin^(-1)(0.67)
Evaluating the inverse sine:
θ ≈ 42.090° (rounded to the third decimal place).
Therefore, the largest angle at which the mass does not slide down the incline is approximately 42.090° (rounded to the third decimal place).