Forces (3)

Question 1

Two particles P and Q of masses 1kg and Mkg respectively are linked by a light inextensible string passing over a smooth pulley as shown.

Particle P is on a rough slope inclined at 20degrees to the horizontal, where the coefficient of friction between P and the slope is is 0.1

Given that P is about to slide down the plane, find the mass of Q at (2 s.f.)

Answer 1

Resolving vertically for Q
F_net = ma
Mg - T = 0, so T = Mg

Resolving perpendicular for P
F_net = ma
R - (1g)cos20degrees = 0
R = (g)cos20degrees

Friction is limiting, so:
F = μR = 0.1 * (g)cos20degrees
Resolving parallel for P
F_net = ma
(1g)sin20degrees - F - T = 0
(1g)sin20degrees - (0.1g)cos20degrees - Mg = 0
sin20degrees - 0.1cos20degrees = M
M = 0.25kg

Question 2

Two particles P and Q of masses 1kg and Mkg respectively are linked by a light inextensible string passing over a smooth pulley as shown.

Particle P is on a rough slope inclined at 20degrees to the horizontal, where the coefficient of friction between P and the slope is is 0.1

When the mass of Q is 1kg, it pulls P up the slope, Find the acceleration of P in this situation

Answer 2

If M = 1kg
Resolving vertically for Q
F_net = ma
1g - T = 1a
T = g - a (1)

Resolving perpendicular for P
R = gcos20degrees
F = μR = (0.1g)cos20degrees

Resolving parallel for P
F_net = ma
T - (1g)sin20degrees - F = 1a
T - (g)sin20degrees - (0.1g)cos20degrees = a

Substituting (1) into (2)
(g - a) - (g)sin20degrees - (0.1g)cos20degrees = a
g - (g)sin20degrees - (0.1g)cos20degrees = 2a
5.527… = 2a
a = 2.76 ms^-2

Question 3

A box of mass 4kg rests on a rough table, connected to an identical box, hanging over the edge of the table, by a light, inextensible string passing over a smooth pulley. The box is held in place by a force, D, acting perpendicular to the table as shown. The coefficient of friction between the box and the table is 0.6

Find the minimum magnitude of D required for the boxes to remain at rest

Answer 3

Resolving vertically for the hanging box
T - 4g = 0
T = 4g

Resolving horizontally for the box on the table
T - F = 0 => F = T = 4g

Resolving vertically for the box on the table
R - D - 4g = 0 => R = D + 4g

F <= μR, so:
4g <= 0.6(D + 4g)
4g <= 0.6D + 2.4g
1.6g <= 0.6D
D >= 26.133... N
So the minimum magnitude of D is 26.1 N

Question 4

A box of mass 4kg rests on a rough table, connected to an identical box, hanging over the edge of the table, by a light, inextensible string passing over a smooth pulley. The box is held in place by a force, D, acting perpendicular to the table as shown. The coefficient of friction between the box and the table is 0.6

The force D is removed. Find the velocity of the boxes 2 seconds later (assuming that the boxes don’t hit the floor or the pulley)

Answer 4

Resolving vertically for the hanging box
F_net = ma
4g - T = 4a => T = 4g - 4a (1)

Resolving vertically for the box on the table
R - 4g = 0
R = 4g

Friction is limiting so
F = μR => F = 0.6 * 4g = 2.4g

Resolving horizontally for the box on the table
F_net = ma
T - F = 4a => T = 2.4g + 4a (2)

Substituting (1) into (2)
4g - 4a = 2.4g + 4a
1.6g = 8a
a = 1.96 ms^-2

u = 0, v = v, a = 1.96, t = 2
v = u + at => v = 3.92 ms^-1

Question 5

Two particles of mass 5kg and 7kg are connected by a light inextensible string passing over a smooth pulley as shown. The coefficient of friction between each particle and the rough surface is 0.15 and neither particle strikes the pulley

Shows that the frictional force acting on each particle can be written in the form Ag cos pheta for values of A and pheta to be found

Answer 5

Resolving perpendicular for 5kg particle
R_1 - (5g)cos30degrees = 0 => R_1 = (5g)cos30degrees

Friction is limiting so:
F_1 = μR_1 => F_1 = 0.15 * (5g)cos30degrees = (0.75g)cos30degrees

Resolving perpendicular for 7kg particle:
R_2 - (7g)cos40degrees = 0 => R_2 = (7g)cos40degrees

Friction is limiting so:
F_2 = μR_2 => F_2 = 0.15 * (7g)cos40degrees = (1.05g)cos40degrees

Question 6

Two particles of mass 5kg and 7kg are connected by a light inextensible string passing over a smooth pulley as shown. The coefficient of friction between each particle and the rough surface is 0.15 and neither particle strikes the pulley

Hence find the acceleration of 5kg particle

Answer 6

Resolving parallel for the 5kg particle
F_net = ma
T - F_1 - (5g)sin30degrees = 5a
T = 5a + (0.75g)cos30degrees + (5g)sin30degrees (1)

Resolving parallel for the 7kg particle
F_net = ma
(7g)sin40degrees - T - F_2 = 7a
T = (7g)sin40degrees - (1.05g)cos40degrees - 7a (2)

Substitute (1) into (2)
12a = g(7sin40degrees - 1.05cos40degrees - 5sin30degrees - 0.75cos30degrees)
a = 9.8(0.5456…) / 12 = 0.4456…
0.446 ms^-2 (3 s.f.)