(a) State the conditions of equilibrium for a number of coplanar parallel forces. (b) A metre rule is found to balance horizontally at the 48 cm mark. When ...
(a) State the conditions of equilibrium for a number of coplanar parallel forces.
(b) A metre rule is found to balance horizontally at the 48 cm mark. When a body of mass 60 g is suspended at the 6 cm mark, the balance point is found to be at the 30 cm mark. Calculate the;
(i) mass of the metre rule;
(ii) distance of the balance point from the zero end, if the body were moved to the 13 cm mark.
(c) a man pulls up a box of mass 70 kg using an inclined plane of effective length 5 m unto a platform 2.5 m high at a uniform speed. If the frictional force between the box and the plane is 1000 N;
(i) draw a diagram to illustrate all the forces acting on the box while in motion;
(ii) calculate the I. minimum effort applied in pulling up the box; II. velocity ratio of the plane, if it is inclined at 30° to the horizontal; Ill. force ratio of the plane.
(a) Conditions of equilibrium for coplanar parallel forces
A number of coplanar parallel forces are in equilibrium when:
The algebraic sum of the forces is zero, i.e. the sum of the forces acting in one direction equals the sum of the forces acting in the opposite direction (the resultant force is zero).
The algebraic sum of the moments of the forces about any point is zero, i.e. the sum of the clockwise moments about the point equals the sum of the anticlockwise moments about that same point (the principle of moments).
(b) Balancing a metre rule
The rule balances by itself at the \(48\,\text{cm}\) mark, so the whole weight of the rule acts at the \(48\,\text{cm}\) mark.
(i) Mass of the metre rule
With the \(60\,\text{g}\) body hung at the \(6\,\text{cm}\) mark, the new balance point (fulcrum) is at the \(30\,\text{cm}\) mark. The body sits on one side of the fulcrum and the weight of the rule acts on the other side. Taking moments about the \(30\,\text{cm}\) fulcrum:
(ii) New balance point with the body at the 13 cm mark
Let the new balance point be at the \(x\,\text{cm}\) mark. The body (\(60\,\text{g}\)) now acts at \(13\,\text{cm}\) and the rule's weight (\(80\,\text{g}\)) still acts at \(48\,\text{cm}\). Taking moments about the fulcrum at \(x\):
Box mass \(=70\,\text{kg}\), so its weight \(W=mg=70\times10=700\,\text{N}\); length of plane \(L=5\,\text{m}\); height of platform \(h=2.5\,\text{m}\); frictional force \(F=1000\,\text{N}\); angle of incline \(\theta=30^{\circ}\).
(i) Diagram of the forces acting on the box while in motion
The four forces acting on the box are: its weight \(W\) acting vertically downwards; the normal (reaction) force \(N\) acting perpendicular to the surface of the plane; the effort \(E\) applied up along the plane; and the frictional force \(F\) acting down along the plane, opposing the upward motion.
Free-body diagram of the box on the 30° inclined plane: weight W acts vertically down, normal reaction N perpendicular to the plane, effort E up the plane, and friction F down the plane opposing motion.
(ii) Calculations
I. Minimum effort applied in pulling up the box
Since the box moves up at uniform (constant) speed, the effort must balance the component of the weight along the plane together with the friction acting down the plane:
(a) Conditions of equilibrium for coplanar parallel forces
A number of coplanar parallel forces are in equilibrium when:
The algebraic sum of the forces is zero, i.e. the sum of the forces acting in one direction equals the sum of the forces acting in the opposite direction (the resultant force is zero).
The algebraic sum of the moments of the forces about any point is zero, i.e. the sum of the clockwise moments about the point equals the sum of the anticlockwise moments about that same point (the principle of moments).
(b) Balancing a metre rule
The rule balances by itself at the \(48\,\text{cm}\) mark, so the whole weight of the rule acts at the \(48\,\text{cm}\) mark.
(i) Mass of the metre rule
With the \(60\,\text{g}\) body hung at the \(6\,\text{cm}\) mark, the new balance point (fulcrum) is at the \(30\,\text{cm}\) mark. The body sits on one side of the fulcrum and the weight of the rule acts on the other side. Taking moments about the \(30\,\text{cm}\) fulcrum:
(ii) New balance point with the body at the 13 cm mark
Let the new balance point be at the \(x\,\text{cm}\) mark. The body (\(60\,\text{g}\)) now acts at \(13\,\text{cm}\) and the rule's weight (\(80\,\text{g}\)) still acts at \(48\,\text{cm}\). Taking moments about the fulcrum at \(x\):
Box mass \(=70\,\text{kg}\), so its weight \(W=mg=70\times10=700\,\text{N}\); length of plane \(L=5\,\text{m}\); height of platform \(h=2.5\,\text{m}\); frictional force \(F=1000\,\text{N}\); angle of incline \(\theta=30^{\circ}\).
(i) Diagram of the forces acting on the box while in motion
The four forces acting on the box are: its weight \(W\) acting vertically downwards; the normal (reaction) force \(N\) acting perpendicular to the surface of the plane; the effort \(E\) applied up along the plane; and the frictional force \(F\) acting down along the plane, opposing the upward motion.
Free-body diagram of the box on the 30° inclined plane: weight W acts vertically down, normal reaction N perpendicular to the plane, effort E up the plane, and friction F down the plane opposing motion.
(ii) Calculations
I. Minimum effort applied in pulling up the box
Since the box moves up at uniform (constant) speed, the effort must balance the component of the weight along the plane together with the friction acting down the plane:
