Figure 6-24 shows an initially stationary block of mass m on a floor. A force of magnitude 0.540mg is then applied at upward angle 16 degrees. What is the magnitude of the acceleration of the block across the floor if (a)Us=.6 and Uk=.5 and (b) Us=.4 and Uk=.3?
Is the 0.540mg 0.540 milligrams or 0.54m * g? I will assume the latter.
For case (a), the maximum possible static friction force is
Fs = (m*g cos16* - 0.54m*g sin16)*Us
= 0.487 m g
The horizontal component of applied force is 0.540m*g*cos 16 = 0.519 m*g. Since this exceeds the maximum possible static friction force, the block slips and accelerates at a rate determined by
kinetic friction.
The friction force in the horizontal direction must be recomputed and subtracted from the applied force component in that direction.
For case (b), proceed similarly
Well, well, well! Looks like we've got ourselves a block on the move! Let's unleash some humor and tackle this together!
(a) Us=.6 and Uk=.5:
To find the magnitude of the acceleration, we need to evaluate the net force acting on the block. Since the force applied is at an angle, we first need to find the component of the force parallel to the floor.
Now, I don't want to "angle" you, but we need a little trigonometry here. The component of the force parallel to the floor is given by F_parallel = F_applied * sinθ, where θ is the angle of 16 degrees.
Once we have that, we can find the net force as the difference between the applied force and the force of friction, which is μk * m * g, where μk is the coefficient of kinetic friction.
Finally, we divide the net force by the mass of the block to find the magnitude of acceleration.
(b) Us=.4 and Uk=.3:
Oh, the coefficients of friction have changed, eh? No worries, let's repeat the process!
Use the same formulas as before, but this time substitute the new values of μs and μk. Calculate the component of the force parallel to the floor, find the net force, and divide by the mass of the block.
Voila! We've computed the magnitude of acceleration for both cases. Now, go ahead and crunch those numbers. Remember, laughter is the best mathemati-trick to solve any problem!
To find the magnitude of the acceleration of the block across the floor, we need to consider the forces acting on the block and apply Newton's second law of motion.
Let's start by analyzing the forces acting on the block:
1. Weight (mg): This force acts vertically downward and has a magnitude of mg, where m is the mass of the block and g is the acceleration due to gravity.
2. Applied force (F): This force is applied at an upward angle of 16 degrees. Its magnitude is given by 0.540mg.
3. Normal force (N): This force occurs perpendicular to the floor and is equal in magnitude to the weight (mg) since the block is initially stationary.
4. Friction force (Ff): This force opposes the applied force and is dependent on the coefficient of static friction (Us) and the normal force (N).
(a) When Us = 0.6 and Uk = 0.5:
To determine if the block moves or remains stationary, we need to determine the maximum static friction force (Fs_max) and compare it to the applied force.
The maximum static friction force can be calculated using the formula: Fs_max = Us * N.
1. Calculate the normal force (N) acting on the block: N = mg.
2. Calculate Fs_max: Fs_max = Us * N.
3. Compare Fs_max to the applied force (F):
- If Fs_max >= F, the block does not move and the acceleration is 0.
- If Fs_max < F, the block moves and the magnitude of the acceleration can be calculated using: a = (F - Fs_max) / m.
(b) When Us = 0.4 and Uk = 0.3:
To determine if the block moves or remains stationary, we need to determine the maximum static friction force (Fs_max) and the kinetic friction force (Fk). We will compare them to the applied force (F) for both cases.
The maximum static friction force (Fs_max) can be calculated using the formula: Fs_max = Us * N.
The kinetic friction force (Fk) can be calculated using the formula: Fk = Uk * N.
1. Calculate the normal force (N) acting on the block: N = mg.
2. Calculate Fs_max: Fs_max = Us * N.
3. Calculate Fk: Fk = Uk * N.
4. Compare Fs_max and Fk to the applied force (F):
- If Fs_max >= F, the block does not move, and the acceleration is 0.
- If Fs_max < F and Fk >= F, the block moves, and the magnitude of the acceleration can be calculated using: a = (F - Fs_max) / m.
- If Fk < F, the block moves, and the magnitude of the acceleration can be calculated using: a = Fk / m.
These steps outline the process to calculate the magnitude of the acceleration for both cases. Remember to use the appropriate values for the given coefficients of friction, mass, and angle of application.
To find the magnitude of the acceleration of the block across the floor in both cases, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Let's break down the problem into two components: the force parallel to the floor (F_parallel) and the force perpendicular to the floor (F_perpendicular).
(a) Us = 0.6 and Uk = 0.5:
1. Determine the force perpendicular to the floor (F_perpendicular):
F_perpendicular = F_applied * cos(theta)
F_perpendicular = (0.540mg) * cos(16 degrees)
2. Determine the force parallel to the floor (F_parallel):
F_parallel = F_applied * sin(theta)
F_parallel = (0.540mg) * sin(16 degrees)
3. Determine the force of friction (F_friction):
F_friction = Us * F_perpendicular
F_friction = (0.6) * F_perpendicular
4. Determine the acceleration of the block across the floor:
F_parallel - F_friction = m * a
(0.540mg * sin(16 degrees)) - (0.6 * 0.540mg * cos(16 degrees)) = m * a
(b) Us = 0.4 and Uk = 0.3:
Follow the same steps as in case (a), but substitute the values of Us and Uk with 0.4 and 0.3, respectively.
Once you calculate the values for F_perpendicular, F_parallel, F_friction, and a for both cases, you will have the magnitude of the acceleration of the block across the floor.