A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are us=0.84 and uk=0.71, respectively.
The acceleration of gravity is 9.8 m/s^2.
What is the frictional force acting on the 29 kg mass?
Answer in units of N.
It matters what the incline is.
To find the frictional force acting on the 29 kg mass, we need to determine whether the block is in a state of static or kinetic friction.
If the block is at rest, the static frictional force will be acting on it. The static frictional force can be calculated using the formula:
Fs ≤ us * N
where Fs is the static frictional force, us is the coefficient of static friction, and N is the normal force.
The normal force can be calculated using the formula:
N = m * g
where m is the mass of the block (29 kg) and g is the acceleration due to gravity (9.8 m/s^2).
Substituting the given values:
N = 29 kg * 9.8 m/s^2
N = 284.2 N
Now, substituting the values of us and N into the static friction equation:
Fs ≤ 0.84 * 284.2 N
Thus, the maximum static frictional force is:
Fs ≤ 238.728 N
Since the block is at rest, the frictional force will be equal to the maximum static frictional force. Therefore, the frictional force acting on the 29 kg mass is approximately 238.728 N.