Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 2 (Q.No. 19)
19.
A block of mass m1, placed on an inclined smooth plane is connected by a light string passing over a smooth pulley to mass m2, which moves vertically downwards as shown in the below figure. The tension in the string is
m1/m2
m1.g sin α
m1.m2/m1+m2
Discussion:
13 comments Page 1 of 2.
Gaurav said:
1 decade ago
If angle = 90 then formula becomes 2m1m2g/m1+m2.
Sanju said:
1 decade ago
Anyone explain this?
Seshu said:
10 years ago
Can any body explain how it obtains?
Sunil said:
9 years ago
I don't understand. Please, anyone solve this problem.
Benny said:
9 years ago
What is g?
Describe the answer.
Describe the answer.
Mohamed Irfan said:
9 years ago
Say it move with acceleration a then using the newtons law,
F = ma.
for mass m1 : m2g-T=m2a ------> 1.
for mass m1 : T-m1gsinx=m2a -----> 2.
If you solve this above equitation you will get T.
F = ma.
for mass m1 : m2g-T=m2a ------> 1.
for mass m1 : T-m1gsinx=m2a -----> 2.
If you solve this above equitation you will get T.
Zaahir said:
9 years ago
To get accelaration you all can directly apply F = ma to total system. Then you can consider one system to get the tension as well.
Satar said:
8 years ago
According to the 2nd law of newton,
For all the system;
f(net)=(m1+m2)a=m2g-m1gsin&alpa;
a=(m2g-m1gsin@)/(m1+m2).
For the mass no.2 from the free body,
F(net)=m2a.
m2-T=m2(m2g-m1gsin@)/(m1+m2).
For all the system;
f(net)=(m1+m2)a=m2g-m1gsin&alpa;
a=(m2g-m1gsin@)/(m1+m2).
For the mass no.2 from the free body,
F(net)=m2a.
m2-T=m2(m2g-m1gsin@)/(m1+m2).
Chittaranjan Mahananda said:
8 years ago
Nice. Thank you all.
Ram said:
6 years ago
Please explain.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers