What force would you have to exert on a 326-N trunk up a 19.0° inclined plane so that it would slide down the plane with a constant velocity? What would be the direction of the force? (The coefficient of friction between the plane and the trunk is 0.328.)
The force of friction is muFn, which is muW. F applied = F parallel (+)(-) F of friction. The trunk is going down the plane so you would subtract?? Am I thinking in the correct direction??
First, figure out whether the trunk would slide down the hill on its own if no external force were applied. If so, then a force UP the hill would have to be applied to have it move at constant velocity.
The friction force when moving is UP the hill, and is
(0.328)(326 cos 18) = 101.1 N
The gravity force component DOWN the hill is 326 sin 19 = 106.1. The difference of 5.0 N must be applied UP the hill (along the slope direction) for contant-velocity sliding.
thank you sooooooooooooooo uch for the honmework help mannnnn @drwls
Thank you for the help @drwls!!!
Well, well, well! It seems like you're in a bit of a pickle, my friend. So, you're trying to figure out the force needed to make that trunk slide down the inclined plane with a constant velocity, huh? Let's break it down.
First off, we have the force of gravity pulling the trunk downwards, which can be calculated as m*g, where m is the mass and g is the acceleration due to gravity. In this case, since we know the weight (W) of the trunk is 326 N, we can find the mass using the equation W = m*g.
Next, we need to consider the force of friction acting against the motion of the trunk. The coefficient of friction (mu) between the plane and the trunk is given as 0.328. The force of friction (Ffriction) can be calculated as mu times the normal force (Fn), which is perpendicular to the surface of the inclined plane.
Since the trunk is on the inclined plane, we need to find the component of the weight that acts parallel to the plane, which is W*sin(theta), where theta is the angle of inclination. This is the force (Fparallel) that needs to be overcome in order to slide the trunk down the plane.
Finally, we have the applied force (Fapplied) that you need to figure out. According to Newton's second law, Fapplied = Fparallel - Ffriction.
Now, considering the direction of the forces: since the trunk is going down the plane, the applied force and force of friction will have opposite directions. So, in order for the trunk to slide down the plane with a constant velocity, the applied force needs to be greater than the force of friction, but in the opposite direction.
I hope this helps you solve your problem, my friend! But remember, if physics gets too tough, you can always join the circus. They're always in need of clowns like me!
Yes, your approach is correct. To determine the force required to slide the trunk down the inclined plane with a constant velocity, you need to consider the component of the force of gravity parallel to the plane, as well as the force of friction opposing the motion.
Step 1: Calculate the component of the force of gravity parallel to the plane:
The force of gravity acting on the trunk can be split into two components: one perpendicular to the plane and one parallel to the plane.
F_parallel = m * g * sin(theta)
where m is the mass of the trunk, g is the acceleration due to gravity (9.8 m/s^2), and theta is the angle of inclination (19.0°).
Step 2: Calculate the force of friction:
The force of friction can be determined using the equation:
F_friction = μ * F_normal
where μ is the coefficient of friction and F_normal is the normal force acting on the trunk.
The normal force can be calculated as:
F_normal = m * g * cos(theta)
Therefore,
F_friction = μ * m * g * cos(theta)
Step 3: Calculate the net force required:
Since the trunk is moving with a constant velocity, the net force acting on it must be zero. Therefore, the applied force must be equal to the combination of the parallel component of gravity force and the force of friction:
F_applied = F_parallel + (-F_friction) [Note: The negative sign is because friction opposes motion]
Now, substitute the values and solve for F_applied:
F_applied = (m * g * sin(theta)) - (μ * m * g * cos(theta))
= m * g * (sin(theta) - μ * cos(theta))
Step 4: Calculate the force required to slide the trunk down with a constant velocity:
Substitute the given values:
m = 326 N, theta = 19.0°, μ = 0.328, g = 9.8 m/s^2
F_applied = 326 * 9.8 * (sin(19.0°) - 0.328 * cos(19.0°))
Using a calculator, you can evaluate this expression to find the force required.
Regarding the direction of the force, since the trunk is moving down the inclined plane, the force applied should be in the opposite direction of the motion. Therefore, the force should be exerted upwards along the incline.