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June 2011 p43 q5
3978
A small block of mass 1.25 kg is on a horizontal surface. Three horizontal forces, with magnitudes and directions as shown in the diagram, are applied to the block. The angle \(\theta\) is such that \(\cos \theta = 0.28\) and \(\sin \theta = 0.96\). A horizontal frictional force also acts on the block, and the block is in equilibrium.
Show that the magnitude of the frictional force is 7.5 N and state the direction of this force.
Given that the block is in limiting equilibrium, find the coefficient of friction between the block and the surface.
The force of magnitude 6.1 N is now replaced by a force of magnitude 8.6 N acting in the same direction, and the block begins to move. Find the magnitude and direction of the acceleration of the block.
Solution
(i) Resolve forces in the x-direction:
\(F_x = 6.1 - 5 \times 0.28 = 0\)
Resolve forces in the y-direction:
\(F_y = 4.8 - 5 \times 0.96 = 0\)
The frictional force acts parallel to the x-axis and to the right. Therefore, \(F_y = 0 \Rightarrow F = F_x\). The frictional force has magnitude 7.5 N.
(ii) Using \(F = \mu R\) and \(R = mg\), we have:
\(\mu = \frac{7.5}{1.25 \times 10}\)
The coefficient of friction is 0.6.
(iii) Applying Newton's second law:
\([7.5 - 8.6 - 1.4 = 1.25a \Rightarrow a = -2]\)
The magnitude of acceleration is 2 m/s2 and the direction is parallel to the x-axis and to the left.