An atwood machine consists of two masses, ma = 7.0 kg and mb = 8.0 kg , connected by a cord that passes over a pulley free to rotate about a fixed axis. the pulley is a solid cylinder of radius r0 = 0.40 m and mass 0.70 kg .

Explanation:

Given Information

Block A mass, mA = 7 Kg

Block B mass, mB=8 Kg

Pulley mass, M=0.7 kg

Pulley Radius, R=0.4 m

I=moment of inertia=MR^2/2 =0.35R^2

Looking for

Acceleration of both blocks.

Percentage error of the two cases (the pulley’s moment of inertia is ignored and the pulley’s moment of inertia is not ignored).

1) mass acceleration a = g(mB – mA)/(mB + mA + I/R^2) = 9.8*1/(8+7+0.35) = 0.638 (m/s2), cylinder acceleration ε = a/R = 1.6m(s-2)

2)

In case the moment of inertia of the pulley is ignored mass acceleration a1 = 9.8*1/(8+7) = 0.653 (m/s2), % error = 100(aignored-a)/aignored = 2.30%.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

at right angles to each other