Limiting equilibrium both directions

6 A block of mass 5 kg is placed on a plane inclined at \(30 ^ { \circ }\) to the horizontal. The coefficient of friction between the block and the plane is \(\mu\).

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{fcc3d739-5c36-48ad-9c34-f69b28a06dba-10_424_709_392_760} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
   \end{figure} When a force of magnitude 40 N is applied to the block, acting up the plane parallel to a line of greatest slope, the block begins to slide up the plane (see Fig. 6.1). Show that \(\mu < \frac { 1 } { 5 } \sqrt { 3 }\).
2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{fcc3d739-5c36-48ad-9c34-f69b28a06dba-11_422_727_264_749} \captionsetup{labelformat=empty} \caption{Fig. 6.2}
   \end{figure} When a force of magnitude 40 N is applied horizontally, in a vertical plane containing a line of greatest slope, the block does not move (see Fig. 6.2). Show that, correct to 3 decimal places, the least possible value of \(\mu\) is 0.152 .

4 A block of mass 11 kg is at rest on a rough plane inclined at \(30 ^ { \circ }\) to the horizontal. A force acts on the block in a direction up the plane parallel to a line of greatest slope. When the magnitude of the force is \(2 X \mathrm {~N}\) the block is on the point of sliding down the plane, and when the magnitude of the force is \(9 X \mathrm {~N}\) the block is on the point of sliding up the plane. Find

1. the value of \(X\),
2. the coefficient of friction between the block and the plane.

6 A block of weight 6.1 N is at rest on a plane inclined at angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac { 11 } { 60 }\). The coefficient of friction between the block and the plane is \(\mu\). A force of magnitude 5.9 N acting parallel to a line of greatest slope is applied to the block.

1. When the force acts up the plane (see Fig. 1) the block remains at rest. Show that \(\mu \geqslant \frac { 4 } { 5 }\).
2. When the force acts down the plane (see Fig. 2) the block slides downwards. Show that \(\mu < \frac { 7 } { 6 }\).
3. Given that the acceleration of the block is \(1.7 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) when the force acts down the plane, find the value of \(\mu\).

5 \begin{figure}[h] \end{figure} A force, whose direction is upwards parallel to a line of greatest slope of a plane inclined at \(35 ^ { \circ }\) to the horizontal, acts on a box of mass 15 kg which is at rest on the plane. The normal component of the contact force on the box has magnitude \(R\) newtons (see Fig. 1).

1. Show that \(R = 123\), correct to 3 significant figures. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{fcd2b219-d9b4-4972-b8fe-25cf543b9054-3_369_1045_1247_555} \captionsetup{labelformat=empty} \caption{Fig. 2}
   \end{figure} When the force parallel to the plane acting on the box has magnitude \(X\) newtons the box is about to move down the plane, and when this force has magnitude \(5 X\) newtons the box is about to move up the plane (see Fig. 2).
2. Find the value of \(X\) and the coefficient of friction between the box and the plane.
3. A particle \(P\) of mass 1.2 kg is released from rest at the top of a slope and starts to move. The slope has length 4 m and is inclined at \(25 ^ { \circ }\) to the horizontal. The coefficient of friction between \(P\) and the slope is \(\frac { 1 } { 4 }\). Find
   1. the frictional component of the contact force on \(P\),
   2. the acceleration of \(P\),
   3. the speed with which \(P\) reaches the bottom of the slope.
   4. After reaching the bottom of the slope, \(P\) moves freely under gravity and subsequently hits a horizontal floor which is 3 m below the bottom of the slope.
      (a) Find the loss in gravitational potential energy of \(P\) during its motion from the bottom of the slope until it hits the floor.
      (b) Find the speed with which \(P\) hits the floor.
      [0pt] [1]
      [0pt] [3] \(7 \quad\) A particle \(P\) starts to move from a point \(O\) and travels in a straight line. At time \(t\) s after \(P\) starts to move its velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), where \(v = 0.12 t - 0.0006 t ^ { 2 }\).
      1. Verify that \(P\) comes to instantaneous rest when \(t = 200\), and find the acceleration with which it starts to return towards \(O\).
      2. Find the maximum speed of \(P\) for \(0 \leqslant t \leqslant 200\).
      3. Find the displacement of \(P\) from \(O\) when \(t = 200\).
      4. Find the value of \(t\) when \(P\) reaches \(O\) again.

3 \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} A block of weight 7.5 N is at rest on a plane which is inclined to the horizontal at angle \(\alpha\), where \(\tan \alpha = \frac { 7 } { 24 }\). The coefficient of friction between the block and the plane is \(\mu\). A force of magnitude 7.2 N acting parallel to a line of greatest slope is applied to the block. When the force acts up the plane (see Fig. 1) the block remains at rest.

1. Show that \(\mu \geqslant \frac { 17 } { 24 }\). When the force acts down the plane (see Fig. 2) the block slides downwards.
2. Show that \(\mu < \frac { 31 } { 24 }\).

5 A particle of mass \(m \mathrm {~kg}\) is resting on a rough plane inclined at \(30 ^ { \circ }\) to the horizontal. A force of magnitude 10 N applied to the particle up a line of greatest slope of the plane is just sufficient to stop the particle sliding down the plane. When a force of 75 N is applied to the particle up a line of greatest slope of the plane, the particle is on the point of sliding up the plane. Find \(m\) and the coefficient of friction between the particle and the plane.

10 A body of mass 20 kg is on a rough plane inclined at angle \(\alpha\) to the horizontal.
The body is held at rest on the plane by the action of a force of magnitude \(P \mathrm {~N}\).
The force is acting up the plane in a direction parallel to a line of greatest slope of the plane.
The coefficient of friction between the body and the plane is \(\mu\).

1. When \(P = 100\), the body is on the point of sliding down the plane. Show that \(g \sin \alpha = g \mu \cos \alpha + 5\).
2. When \(P\) is increased to 150, the body is on the point of sliding up the plane. Use this, and your answer to part (a), to find an expression for \(\alpha\) in terms of \(g\).

\includegraphics{figure_4} A heavy package is held in equilibrium on a slope by a rope. The package is attached to one end of the rope, the other end being held by a man standing at the top of the slope. The package is modelled as a particle of mass 20 kg. The slope is modelled as a rough plane inclined at \(60°\) to the horizontal and the rope as a light inextensible string. The string is assumed to be parallel to a line of greatest slope of the plane, as shown in Figure 4. At the contact between the package and the slope, the coefficient of friction is 0.4.

1. Find the minimum tension in the rope for the package to stay in equilibrium on the slope. [8]

The man now pulls the package up the slope. Given that the package moves at constant speed,

1. find the tension in the rope. [4]
2. State how you have used, in your answer to part (b), the fact that the package moves
   1. up the slope,
   2. at constant speed.
   [2]

A body of mass 20 kg is on a rough plane inclined at angle \(\alpha\) to the horizontal. The body is held at rest on the plane by the action of a force of magnitude \(P\) N. The force is acting up the plane in a direction parallel to a line of greatest slope of the plane. The coefficient of friction between the body and the plane is \(\mu\).

1. When \(P = 100\), the body is on the point of sliding down the plane. Show that \(g \sin \alpha = g\mu \cos \alpha + 5\). [4]
2. When \(P\) is increased to 150, the body is on the point of sliding up the plane. Use this, and your answer to part (a), to find an expression for \(\alpha\) in terms of \(g\). [3]