0.4kg
Q0.7 kg
Two particles P and Q. of masses 0.4kg and 0.7 kg respectively, are attached to the ends of a light
inextensible string. The string passes over a fixed smooth pulley which is attached to the edge of a
rough plane. The coefficient of friction between P and the plane is 0.5. The plane is inclined at an
angle a to the horizontal, where tan a = Particle P lies on the plane and particle Q hangs vertically.
The string between P and the pulley is parallel to a line of greatest slope of the plane (see diagram).
A force of magnitude XN, acting directly down the plane, is applied to P.
(i) Show that the greatest value of X for which P remains stationary is 6.2.
(ii) Given instcad that X =0.8, find the acceleration of P.
Answers
Answer:i) with the information: tan=3/4
We can draw a right angle triangle with base=4, hypotenuse=5, and perpendicular=3. From this information we can derive the following expressions:
Cos=4/5
Sin=3/5
The tension in the string=0.7 multiply 10=7N
Normal reaction force at P=(0.4)(10)(Cos)
=4(4/5)=3.2N
F=(coefficient of frictional force)(Normal reaction force)
=(0.5)(3.2)=1.6N
Now,
F=X+(4)Sin-T
1.6=X+(4)(3/5)-7
X=23/5+1.6
X=6.2N
ii)for particle Q,F=ma
Equation1:
(0.7)(10)-T=0.7a
For particle P
Equation 2:
T-0.8-(0.4)(10)Sin-F
T-0.8-(4)(3/5)-1.6=0.4a
T-24/5=0.4a
By solving these two equations simultaneously we have the value of a=2ms^-2
Explanation:
The value of acceleration is a = 2 m/s^2.
Explanation:
We are given that:
- Mass of particle P = 0.4 Kg
- Mass of particle Q = 0.7 Kg
- Coefficient of friction = 0.5
Solution:
The values of cos and sin are given.
- Cos = 4 / 5
- Sin = 3 / 5
Tension in the string = 0.7 = 0.7 x 10 = 7 N
Normal reaction force at P = (0.4) (10) (Cos)
= 4 (4 / 5) = 3.2 N
F = [ Coefficient of frictional force x Normal reaction force ]
= [ 0.5 x 3.2] = 1.6 N
Now,
F = x + (4) Sin - T
1.6 = x + (4) (3 / 5) - 7
x = 23 / 5 + 1.6 = 4.6 + 1.6 = 6.2 N
F = ma for particle Q using equation 1.
(0.7) (10) - T = 0.7 a
For particle P using equation 2.
T - 0.8 - (0.4) (10) Sin - F
T - 0.8 - (4) (3 / 5) - 1.6 = 0.4 a
T - 24 / 5 = 0.4 a
a = 2m / s^2
Thus the value of acceleration is a = 2 m/s^2.