I assume that Theta is up and down from horizontal
calling theta = A
d sin A /dA = cos A
d sin A/dt = d sin A/dA * dA/dt
so
d sin A /dt = cos A (dA/dt )
similarly
d cos A/dt = -sin A dA/dt
That settled, let's look at the problem
Force = m * a
a = F/7
da/dt = (1/7) dF/dt
F in direction of motion = 20 cos A
d F/dt =20 d/dt(cosA)
but we already know what that is
dF/dt = 20 (-sin A) dA/dt
A = 28 deg = .489 radian
sin A = .469
dA/dt = 3*10^-2 (pi/180) = 5.24*10^-5 rads/s
so
dF/dt = 20 (-.489)(5.24*10^-5)
da/dt =(1/7)dF/dt = (20/7)(-.489)(5.24*10^-5)
The figure shows a 7 kg block being pulled along a frictionless floor by a cord that applies a force of constant magnitude 20 N but with an angle θ(t) that varies with time. When angle θ = 29°, at what rate is the block's acceleration changing if (a)θ(t) = (3 × 10-2 deg/s)t and (b)θ(t) = (-3 × 10-2 deg/s)t ? (Hint: Switch to radians.)
2 answers
Net force=mass*acceleration
20*cosTheta=mass* acceleratin
take the deriviative...
-20SinTheta dTheta/dt=mass*d(acceleration)/dt
yes, dTheta/dt is in radians/sec, so convert it. solve for d(acceleration)/dt and the units will be m/sec^3
20*cosTheta=mass* acceleratin
take the deriviative...
-20SinTheta dTheta/dt=mass*d(acceleration)/dt
yes, dTheta/dt is in radians/sec, so convert it. solve for d(acceleration)/dt and the units will be m/sec^3
2 answers