Question

Q24. a) m1 = 5 kg and m2 = 8 kg. A block of mass m1 is on a ramp that is inclined at 20° above the horizontal, It is connected by a string to a block of mass m2 that hangs over the top edge of the ramp. The coefficient of kinetic friction between the incline and the m1 block is 0.22. What is the acceleration of the masses and the tension in the string? (3 marks) b) if m1 = 8 kg, and the coefficient of static friction between m1 and the incline is 0.300, find the maximum value of m2 that will not make m1 slide up the incline (ie. when static friction is maximum) (2 marks)​

Answer 1

a) To find the acceleration of the masses and the tension in the string, we need to consider the forces acting on the system.

1. Forces on m1 (block on the ramp):

• - Weight (mg1) acting vertically downward.
• - Normal force (N1) perpendicular to the ramp.
• - Frictional force (f) opposing the motion.

The component of the weight acting parallel to the incline is
`\displaystyle\sf mg_(1)\sin(20°)`, and the component perpendicular to the incline is
`\displaystyle\sf mg_(1)\cos(20°)`.

The normal force
`\displaystyle\sf N_(1)` is equal in magnitude and opposite in direction to the perpendicular component of the weight, so
`\displaystyle\sf N_(1)=mg_(1)\cos(20°)`.

The frictional force
`\displaystyle\sf f` is given by
`\displaystyle\sf f=\mu N_(1)`, where
`\displaystyle\sf \mu` is the coefficient of kinetic friction.

The net force acting on m1 is the force parallel to the incline, which is the difference between the weight component and the frictional force:

`\displaystyle\sf F_{\text{{net}}}=(mg_(1)\sin(20°))-f`.

Using Newton's second law
`\displaystyle\sf F_{\text{{net}}}=m_(1)a`, where
`\displaystyle\sf a` is the acceleration of the masses, we can find the acceleration:

`\displaystyle\sf m_(1)a=(mg_(1)\sin(20°))-f`.

Now, let's consider the forces acting on m2 (hanging block):

• - Weight (mg2) acting vertically downward.
• - Tension (T) in the string, which is acting vertically upward.

Since the two blocks are connected by the string, the tension in the string is the same for both blocks, so
`\displaystyle\sf T` is the tension in the string.

Using Newton's second law for m2, we have:

`\displaystyle\sf m_(2)g_(2)-T=m_(2)a`.

Now we have two equations with two unknowns (a and T). Solving these equations simultaneously will give us the values for acceleration and tension.

b) To find the maximum value of m2 that will not make m1 slide up the incline, we need to consider the forces and the maximum static friction.

When m1 is at the verge of sliding up the incline, the force of static friction
`\displaystyle\sf f_{\text{{static}}}` is at its maximum and is given by
`\displaystyle\sf f_{\text{{static}}}=\mu_{\text{{static}}}N_(1)`, where
`\displaystyle\sf \mu_{\text{{static}}}` is the coefficient of static friction.

The force of static friction
`\displaystyle\sf f_{\text{{static}}}` must be equal to or greater than the component of the weight of m1 parallel to the incline, which is
`\displaystyle\sf mg_(1)\sin(20°)`.

So we have:

`\displaystyle\sf f_{\text{{static}}}=\mu_{\text{{static}}}N_(1)\geq mg_(1)\sin(20°)`.

Using the expression for
`\displaystyle\sf N_(1)` from part a, we can substitute and solve for the maximum value of m2.