PHYS 2210 (Fall 2018) Homework 4: Mechanics and Motion Problems

This document provides detailed solutions to the PHYS 2210 Homework 4 assignment, covering various mechanics problems. The solutions address topics such as trajectory, forces, friction, and pulley systems. Question 1 examines the trajectory of a rocket, while Question 2 analyzes the forces acting on a box on a conveyor belt, including static and kinetic friction. Question 3 explores the forces in single and double pulley systems. Question 4 focuses on an inclined plane, calculating acceleration and normal force. Question 5 deals with forces, friction, and angles, deriving equations for normal force and kinetic friction. Finally, Question 6 analyzes a system with two masses connected by a rope over pulleys, determining acceleration and tension in the rope, and considering the effects of different angles and mass ratios. The solutions utilize Newton's laws of motion, kinematic equations, and free-body diagrams to arrive at the answers.

Q1)
B
C
This is because when the engine is started it will go down but in a deep space, there will be no
force to stop the rocket from moving sideways, therefore when the engine will provide a constant
acceleration then the trajectory of the rocket will be parabolic, and therefore it will be expected
for the trajectory not to be linear
Q2)
a) The conveyor belt is moving with constant speed towards right and a box is placed on it,
therefore when the box is placed the velocity of the box was zero in the direction of the
conveyor belt so there was a relative motion in between conveyor belt and the box, the
force of kinetic friction will act on direction of motion of belt of belt which accelerates
the box in the direction of motion of belt and the box will begin to gain velocity in the
direction of motion of belt.
At the time when velocity of belt and box becomes equal in the right direction then the
box sits on the conveyor belt without slipping and force of static friction keeps the box
fix at one place on the belt
b) Immediately after the box is placed on the belt, box moves to the left side, on the box the
weight of the box acts downwards, normal force on the box acts upward and force up
kinetic friction fk acts in right direction
Yes there is frictional force that acts on the right direction
c) At this stage, weight of the acts downwards, normal force equals to the weight of box acts
upwards a force of static friction act on the box in the right direction.

Q3)
a) Fnet = ma
m = mass of the object
a = acceleration of the object
For a single pulley system, the tension will be acting upwards directions while the weight
of the piano will be acting in the downward direction.
The net force will be equal to
Fnet = T – mg
T = tension in the rope
Mg = mass of the piano
At the point one holds the piano , a= 0 m/s2
Substituting 0 m/s2 for a in the equation Fnet = ma
Fnet = m(0 m/s2)
= 0
Substitute 0 for Fnet = T – mg
0 = T – mg
T = mg
b) For double pulley system
Tension in the rope is acting in upward directions and weight of the piano of the piano in
the downward direction
The net force acting on the piano – rope system
Fnet = 2T – mg
Substituting for Fnet in the equation above
0 = 2T – mg
T = mg/2

Q4)
Fx = max => mgsinƟ – friction = ma
 Mgsin300 = may
Fy = may
Mgcos300 – N = may
Constraints
Friction = 0
ay = 0
mgsin300 = max
ax = gsin300= 9.8 * ½ = 4.9 m/s2
Mgcos300 = - N = 0
Mgcos300 = N
N = 10 * 9.8 * 0.866
N = 84.87 kg/s2
N = 84.87 N
Therefore, acceleration n of block = 4.9 m/s2 and Normal force acting on box – 84.87 kgm/s2
(b) given d = 10 m
Initial velocity , u = o m/s
Acceleration, a = gsinƟ = 4.9 m/s2
Now consider the equation
d = ut + ½ at2
10 = 0 + ½ *4.9 * t2

10 = 2.45 * t2
t2 = 10/2.45 = 4.08
t = √4.08 2.019 = 2.02
t = 2.02 sec
therefore, time taken by box to get to bottom of the incline is 2.02 sec
Q5)
N F*sinƟ
F
Ɵ F*cosƟ
fs
The equation as per the diagram above
N + F*sinƟ = mg
N = mg – F*sinƟ
(b)
F*cosƟ – fs = ma
Fs = F*cosƟ – ma
fs = μk * N = F*cosƟ – ma
μk = F∗cosƟ – ma
mg−F∗sinƟ
c) If Ɵ = 0
N = mg

μk = F – ma
mg
Q6
a)
b) Assuming acceleration is a1 and a2
M1gsinƟ1 – T = m1 * a1
T – M2gsinƟ2 = m2 * a2
c) Constrains
a1 = a2
d) Presenting acceleration in the form a = F/m
a = m1∗gsinƟ 1−m2∗gsinƟ 2
m 1+ m2
e) When mass m1 is greater than the load m2 , the load m1 will pull m2 towards its direction,
and the value sign of tension will be negative indicating the tension will be acting on
opposite direction
f) T = m2*a1 + m2*g*sinƟ2
= m2*[ m 1∗g∗sinƟ 1−m2∗gsinƟ 2
m1+m 2 ] +¿m2*gsinƟ2
= m2∗m1∗gsinƟ 1−m22∗gsinƟ 2+m1∗m2∗gsinƟ 2+m22∗gsinƟ 2
m1+ m2
T = m1∗m2( sinƟ 1+ sinƟ 2)
mi+m2