Physics – Chapter 4Pulleys

St. Augustine Preparatory SchoolOctober 19, 2015

Pulleys

Solving Pulley Questions

• To solve pulley questions, both the force of gravity (Fg) and the net force (Fnet) are going to be important.

In the situation on theright, which mass do you think will drop and whichone will be pulled up? Why?

Pulley Question Number 1Given two masses, 5kg and 3kg, calculate the net force of the system, as well as the acceleration of the system. Assume that the rope has no mass and the pulley is frictionless.

Solution

a. Net Force: Fnet = Fg,m2 – Fg,m1

Fnet = m2g – m1gFnet = (5.00kg)(-9.81m/s2) – (3.00kg)(-9.81m/s2)Fnet = -19.62 N = -19.6 N

b. Acceleration of the system (using Fnet = ma)a = Fnet/m a = (-19.6 N) / (5.00kg + 3.00kg)a = -2.45 m/s2

Pulley Question Number 2

A 10.0g mass is tied to a string. The string is attached to a 500.0g mass and stretched over a pulley, leaving the 10.0g mass suspended above the floor. Determine the time it will take the 10.0g mass to fall a distance of 1.50 meters if starting from rest and the surface is frictionless. (hint: frictionless therefore Fnet = Fg)

SolutionStep 1: Calculate Fnet Fnet = mg = (0.010kg)(-9.81m/s2) = -0.0981 N

Step 2: Calculate the acceleration of the systema = Fnet/m = -0.0981N/(.500kg + 0.010kg) a = -0.192..m/s2

Step 3: Calculate time to fall 1.50m using kinematics formula (d = vit + ½ at2, assuming vi = 0.0m/s)d = vit + ½ at2 can be rearranged and solved as

Try on your own…

• Consider the two-body system shown below. There is a 0.250-kg object accelerating across a rough surface. The sliding object is attached by a string to a 0.100 kg object which is suspended over a pulley. The coefficient of kinetic friction is 0.183. Calculate the acceleration of the boxes.

(hint: Fnet = Fg – Ff)

Solution:1) Find Fg for the hanging mass

Fg = mg = (0.100kg)(-9.81m/s2) = -0.981N2) Find Ff for the block on the table. This block will oppose the motion of the falling block. Remember, FN will be equal and opposite to the force of gravity

Fk = uFN = (0.183)(9.81m/s2)(0.250kg) = 0.4488…N3) Find the net force Fnet = Fg + Fk = -0.981N + 0.4488..N = -0.53219..N4) Find the acceleration using the mass of both boxesa = Fnet / m = -0.53219..N / 0.350kg = -1.52055 m/s2

a = -1.52 m/s2

Applying both pulley’s and inclined planes

Two boxes are connected by a string that runs through a frictionless pulley. If M2 has a mass of 15kg, M1 has a mass of 9.0kg, the coefficient of static friction is 0.150, and the inclined plane has an angle of 20 degrees, calculate the acceleration of the boxes.

SolutionStep 1: Calculate Fg of the hanging boxFg = mg = (15kg)(-9.81m/s2) = -147.15NStep 2: Calculate Fk for the box on the inclined plane (this requires us to calculate the normal force, which requires us to calculate the force of gravity perpendicular. Note: I used + for perpendicular, since I was unable to find an upside down T)a. Fg+ = Fgcos(θ) = (9.0kg)(-9.81m/s2)cos(20°) = -82.965…N

From this, we now know that the normal force is 82.965…Nb. Fk = uFN = (0.150)(82.965…N) = 12.444…NStep 3: Calculate the net forceFnet = -147.15N + 12.444…N = -134.705…NStep 4: Calculate the acceleration using the mass of both boxesa = Fnet / m = -134.705..N / (15.0+9.0)kg = -5.6127m/s2

a = -5.6 m/s2

Things to remember…• Net force is found by finding the force of gravity

on the more massive box and subtracting the force of gravity on the less massive box

• The two boxes have to have the same acceleration since they are connected– This means that to calculate the acceleration, you

must use the mass of both boxes – Fnet = (m1 + m2)a, where m1 and m2 are the masses of

the two boxes