A block is at rest on the incline shown in the
figure. The coefficients of static and kinetic
friction are μs = 0.7 and μk = 0.59, respec-
tively.
The acceleration of gravity is 9.8 m/s2
What is the frictional force acting on the
29 kg mass?
Answer in units of N
(part 2 of 3)
What is the largest angle which the incline
can have so that the mass does not slide down
the incline?
Answer in units of �
(part 3 of 3)
What is the acceleration of the block down
the incline if the angle of the incline is 40� ?
Answer in units of m/s2
Wb = m*g = 29kg * 9.8N/kg = 284.2 N. = Wt. of the block.
1. Fp = 284.2*sin A = Force parallel to the incline.
Fn = 284.2*cos A = Normal or Force perpendicular to the incline.
Fs = us*Fn = 0.7*284.2*cos A=199*cos A
Fp-Fs = m*a = m*0 = 0
284.2*sinA - 199*c0sA = 0
284.2sinA = 199cosA
Divide both sides by cosA:
284.2*(sinA/cosA) = 199
Replace sinA/cosA with tanA:
284.2*tanA = 199
tanA = 188/284.2 = 0.70
A = 35o
Fs = 199*cos35 = 163 N. = Force of static friction.
2. A = 35o.
3. Fp = 284.2*sin40 = 182.7 N.
Fn = 284.2*cos40 = 217.7 N.
Fk = uk*Fn = 0.59*217.7 = 128.4 N. = Force of kinetic friction.
a=(Fp-Fk)/m = 182.7-128.4)=1.87 m/s^2
To find the frictional force acting on the 29 kg mass, we need to determine if the block is at rest or in motion on the incline. Since the block is described as being at rest, we know that the static frictional force is acting to oppose any potential motion.
The formula for static friction is given by:
fs ≤ μs * N
where fs is the static frictional force, μs is the coefficient of static friction, and N is the normal force. Since the block is on an incline, the normal force can be calculated by:
N = mg * cos(θ)
where m is the mass of the block and θ is the angle of the incline.
Let's calculate the frictional force:
1. Calculate the normal force:
N = (29 kg) * (9.8 m/s^2) * cos(θ)
2. Calculate the maximum static frictional force:
fs_max = μs * N
The maximum static frictional force represents the maximum force that can be exerted by static friction before the block starts moving. Since the block is at rest, the actual static frictional force will be equal to the maximum static frictional force:
fs = fs_max
Now, let's move on to part 2 of the question.
To find the largest angle at which the mass does not slide down the incline, we need to consider the components of the gravitational force along and perpendicular to the incline.
The gravitational force component along the incline is given by:
Fg_parallel = mg * sin(θ)
The static frictional force acting up the incline opposes the gravitational force along the incline, so it can be written as:
fs = -Fg_parallel
Since the static frictional force is given by:
fs = μs * N
and the normal force N is equal to:
N = mg * cos(θ)
we can substitute these equations into the previous equation to find the largest angle:
μs * mg * cos(θ) = mg * sin(θ)
Simplifying the equation, we get:
μs * cos(θ) = sin(θ)
Now, solving for θ, we can find the largest angle at which the mass does not slide down the incline.
Moving on to part 3 of the question.
To find the acceleration of the block down the incline when the angle is 40 degrees, we use Newton's second law:
ΣF_parallel = m * a
The net force parallel to the incline is given by:
ΣF_parallel = Fg_parallel - fs
Substituting the previous equations, we have:
ΣF_parallel = mg * sin(θ) - μk * mg * cos(θ)
The acceleration can be calculated by rearranging the equation as:
a = (mg * sin(θ) - μk * mg * cos(θ)) / m
Now, you can plug in the values given in the question (μk = 0.59, θ = 40 degrees) to calculate the acceleration.