Statics of a particle

WME01Topic 1.53 questions

Forces in equilibrium, resolving forces, and limiting friction.

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Original practice questions written in the style of the exam.

Question 14 marks
A particle of mass $5\,\text{kg}$ hangs in equilibrium, supported by two light inextensible strings. One string makes an angle of $30^\circ$ with the vertical and the other makes an angle of $60^\circ$ with the vertical, on the opposite side.

Find the tension in each string.
Mark scheme
- M1 Resolve horizontally: $T_1\sin 30^\circ = T_2\sin 60^\circ$ (1)
Question 26 marks
A particle of mass $2\,\text{kg}$ is held in equilibrium on a rough plane inclined at an angle $\alpha$ to the horizontal, where $\tan\alpha = \tfrac{3}{4}$ . The coefficient of friction between the particle and the plane is . The particle is held by a force of magnitude newtons, acting in the vertical plane containing a line of greatest slope.

sin

0∘

M1 Resolve vertically:

T_1\cos 30^\circ + T_2\cos 60^\circ = 5g = 49

(1)

A1 String at

60^\circ

T_2 = 24.5\,\text{N}

(1)

A1 String at

30^\circ

T_1 = 42.4\,\text{N}

(1)

0.4

horizontal

P

Given that the particle is on the point of slipping down the plane, find the value of $P$ .

Mark scheme

B1 $\sin\alpha = 0.6$ , $\cos\alpha = 0.8$ (1)
M1 Resolve perpendicular to the plane: $R = 2g\cos\alpha + P\sin\alpha = 15.68 + 0.6P$ (1)
B1 Friction acts up the plane with $F = 0.4R$ (1)
M1 Resolve parallel to the plane: $P\cos\alpha + F = 2g\sin\alpha$ (1)
M1 Substitute: $0.8P + 0.4(15.68 + 0.6P) = 11.76$ , so $1.04P = 5.488$ (1)
A1 $P = 5.3$ ( $5.28$ ) (1)