Two particles P and Q have masses 1.5 kg and 3 kg respectively. The particles are attached to the ends of a light inextensible string. Particle P is held at rest on a fixed rough horizontal table. The coefficient of friction between P and the table is $ \frac{1}{5} $. The string is parallel to the table and passes over a small smooth light pulley which is fixed at the edge of the table. Particle Q hangs freely at rest vertically below the pulley, as shown in Figure 3. Particle P is released from rest with the string taut and slides along the table. Assuming that P has not reached the pulley, find a) the tension in the string during the motion, b) the magnitude and direction of the resultant force exerted on the pulley by the string.

A-Level Edexcel Maths Mechanics Newtons Second Law: Two particles P and Q have masse

Two particles P and Q have masses 1.5 kg and 3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2016 - Paper 1

Question 8

Two particles P and Q have masses 1.5 kg and 3 kg respectively. The particles are attached to the ends of a light inextensible string. Particle P is held at rest on ... show full transcript

Worked Solution & Example Answer:Two particles P and Q have masses 1.5 kg and 3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2016 - Paper 1

Step 1

a) the tension in the string during the motion

96%

114 rated

Answer

To find the tension in the string during the motion, we can start with the following equations based on the forces acting on the two particles.

For particle P (1.5 kg):
- The net force acting on P as it moves horizontally can be expressed as: $T - f = 1.5a$ where ( f ) is the frictional force and can be calculated using: $f = \mu R = \frac{1}{5} \times 1.5g$ ( R ) is the normal reaction on P, equal to ( 1.5g ).
For particle Q (3 kg):
- The force acting on Q in the vertical direction: $3g - T = 3a$

Here, we can solve these equations simultaneously. First, we'll substitute for ( f ) in the first equation and isolate ( T ).

From the values, we get:

( f = \frac{1}{5} \times 1.5 \times 9.8 = 2.94 , \text{N} )

Substituting this back, we have: $T - 2.94 = 1.5a$

Similarly, from the second equation: $3g - T = 3a$

Now substituting g = 9.8 m/s²:

=> T = 29.4 - 3a $$ By eliminating \( a \) from the equations and solving, we find the value of tension \( T \). Considering the typical accelerations and forces acting, we finally get: - \( T = 1.5g - f \) resulting in \( T = 12 \text{ N} \).

Step 2

b) the magnitude and direction of the resultant force exerted on the pulley by the string

99%

104 rated

Answer

To find the resultant force exerted on the pulley by the string, we first need to consider the tensions in the strings on either side of the pulley:

The tension acting on particle Q is: ( T_Q = T )
The tension acting on particle P remains ( T_P = T )

The resultant force ( R ) on the pulley can be calculated using the Pythagorean theorem due to the angles involved: $R = \sqrt{T^2 + T^2 \cos^2 45^\circ}$ Plugging in the values: $R = \sqrt{(12)^2 + (12) \cdot \frac{\sqrt{2}}{2}} \approx 16.6 \, \text{N}$

The direction of this resultant force will be at an angle ( 45^\circ ) below the horizontal due to the equal tensions from both sides.

Two particles P and Q have masses 1.5 kg and 3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2016 - Paper 1

Worked Solution & Example Answer:Two particles P and Q have masses 1.5 kg and 3 kg respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2016 - Paper 1

a) the tension in the string during the motion

b) the magnitude and direction of the resultant force exerted on the pulley by the string

Join the A-Level students using SimpleStudy...