A cart of mass 380 grams is temporarily held stationary on an incline?

Question

A cart of mass 380 grams is temporarily held stationary on an incline at 3.7 degrees to horizontal.
•A mass of 75.00001 grams is attached to the cart by a string over a pulley located at the bottom of the incline; the mass hangs freely from the string.
•When the cart is released, what will be its acceleration in the direction down the incline?
•Sketch and label a reasonable facsimile of a scaled diagram showing the forces acting on the cart.

Answer 1

I won't sketch it, I'll describe it verbally

looking at a fbd of the cart at the instant of release, it has
gravity pressing it into the incline cos(3.7)*m*g

and gravity pulling it down the incline

sin(3.7)*m*g

and the tension in the string pulling it up the incline

-T (since we care about a don the incline)

since there's no friction and we assume the pulley and string to have no mass, the forces on the cart parallel to the incline are

sin(3.7)*m*g-T
this is equal to m*a

To compute T, conside a fbd of the hanging mass

gravity downward
-m*g
and T upward
under the assumptions I stated, the acceleration will be the same for both objects and the sign is upward

T-m*g=m*a
The mass is .07500001 kg
T=.07500001*(g-a)

Plugging into the first equation

recall
sin(3.7)*m*g-T=m*a
where m=.380 kg
sin(3.7)*.380*g-
.07500001*(g-a)
=.380*a

doing the algebra
g*(sin(3.7)*.380-.07500001)=
a*(.380-.07500001)

using 9.81 for g
the acceleration down is
-1.62 m/s^2

negative because the cart moves up the incline.

to check note that the component of the cart mass is much less than the hanging mass because the angle is so shallow at 3.7 degrees.

j