Question

Question 7 (15 pts): As you are walking past a house you notice a friend is having trouble setting up their Halloween decorations and you decide to help out. A jack-o-lantern of mass m = 50 kg sits at rest upon a rough inclined plane with an angle of 0 = 30°. A rope is connected to the jack-o-lantern and you need to pull the jack-o-lantern up the ramp. The coefficients of static and kinetic friction are 0.5 and 0.4, respectively, find (a) the smallest force required to initially make the jack-o-lantern move; and, (b) the force required to pull the block up the incline at a constant speed, once it is moving? ) 300

Answer 1

Given the mass of the lantern m = 50 Kg

So the weight of the lantern =mg = 50*9.8 = 490N

The weight,since it is on an incline of 30° can be resolved into 2 components.

mg cos 30 acting down perpandicular to the incline and acting as the normal force

So N = mg cos 30 = 490 cos30°

The mg sin 30 component acting down parallel to the incline.

So the force acting downward= 490 sin 30°

a. The smallest force required to move the lantern up = downward force+ force due to static friction

= 490 sin30 + $\mu$ s .N =

Where $\mu$ s is the coefficient of static friction.

F = 245 + 0.5 * 490 cos 30 = 457.18 N

So the minimum force for making the lantern move = 457.18 N

b.

In the condition of the constant speed of the lantern upward,

The force we apply up must be equal to the sum of opposing downward forces( kinetic friction and 490 sin30

F = $\mu$ k .N + 490 sin 30

Where $\mu$ k is the coefficient of kinetic friction.

F = 0.4* 490 cos 30 + 490 sin30

F = 414.74 N

So the force we need to give to pull up the lantern without acceleration= 414.74 N

(Please upvote if helpful)