Two packing crates of masses m1 = 10.0 kg and m2 = 7.00 kg are connected by a light string that passes over a frictionless pulley as in the figure below. The 7.00 kg crate lies on a smooth incline of angle 41.0°. Find the acceleration of the 7.00 kg crate.

Question

owlfred · Answer

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

anonymous · Answer

where's the diagram?

anonymous · Answer

the equations of the motion :
where T is the tension in the rope, the system will have common acceleration  a since the same rope.  

10g- T = 10a..........(1)
T- 7g  =7a...............(2)

Solving both the equations,
a= 3g/17.   ( g= 9.8m/s^2)
a= 1.72m/s^2

hope this will help

anonymous · Answer

$$Let $$'T' be the tension of the string and 'a' be the acceleration.
 
Consider the motion of the mass m1=10 kg,

According to newton's second law,

W1-T=m1a
m1g-T=m1a          (W=mg)                                               
                             _______(1)

Now consider the motion of body of mass m2=7 kg,

According to newton's second law,

T-W2sin41=m2a
T-m2gsin41=m2a    _______(2)

(1)+(2)=
m1g-T+T-m2gsin41=ma+m2a
m1g-m2gsin41=m1a+m2a
g(m1-m2sin41)=a(m1+m2)
a=g(m1-m2sin41)/(m1+m2)
a=9.81(10=7*sin41)/(10+7)
a=3.12m/s^2

Therefore acceleration of the body of mass 7 kg = 3.12 m/s^2