Applications of forces 7B
1 From symmetry the tension in both strings is the same.
()
R ↑
T sin 45° + T sin 45° − 5 g = 0
∴ 2T sin 45° = 5 g
5g
T=
2 sin 45°
49 2
=
2
T = 34.6 N (3 s.f.)
2 a Let the tension in the string be T N
R(←)
T sin30° − 10 = 0
10
∴T =
sin30°
T = 20 N
b R ↑ ()
T cos 30° − mg = 0
mg = 20 cos 30° (since T = 20 N)
20 cos 30°
∴m =
g
10 3
=
g
= 1.8 kg (2 s.f.)
3 Let the tension in the string be T N.
R(→)
8 − T sin θ = 0
∴T sin θ = 8 (1)
()
R ↑
T cos θ − 12 = 0
∴T cos θ = 12 (2)
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 1
3 a Divide equation (1) by equation (2) to eliminate the tension T.
T sinθ 8
=
T cosθ 12
2
∴ tanθ =
3
∴θ = 33.7° (3 s.f.)
b Substitute into equation (1)
T sin 33.7° = 8
8
T=
sin 33.7°
= 14.4 (3 s.f.)
4 Let the tension in the strings be T N and S N as shown in the figure.
R(←)
T cos 60° − S cos 45° = 0
T S
∴ − =0
2 2
∴T = S 2 (1)
R ↑ ()
T sin 60° + S sin 45° − 6 g = 0
3 1
T +S = 6g (2)
2 2
Substitute T = S 2 from (1) into equation (2)
3 1
S 2 × + = 6g
2 2
3 +1
S = 6 g
2
6g 2
S=
( 3 +1 )
= 3g 2 ( )
3 −1
= 30 (2 s.f.)
and T = 6g ( )
3 − 1 = 43 (2 s.f.)
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 2
5 a Let the tension in the string be T and the mass of the bead
be m.
Resolve horizontally first to find T:
R(→)
T cos 30° − T cos 60° − 2 = 0
T (cos 30° − cos 60°) = 2
2
∴T =
cos 30° − cos 60°
4
=
3 −1
=
4 ( 3 +1 )
( )( 3 + 1)
3 −1
4 ( 3 + 1)
=
2
=2 ( )
3 + 1 = 5.46 N (3 s.f.)
b R ↑ ()
T sin 60° + T sin 30° − mg = 0
mg = T (sin 60° + sin 30°)
3 1
m=
2
g
( )
3 + 1
2
+
2
(using T = 2 ( )
3 + 1 from part a)
4+2 3
=
g
= 0.76kg (2 s.f.)
c Modelling the bead as smooth assumes there is no friction between it and the string.
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 3
6 Let the tension in the string be T and the mass of the bead be m.
a Resolve horizontally first to find T.
R(→)
2 − T cos 60° − T cos 30° = 0
T (cos 60° + cos 30°) = 2
2
∴T =
cos 60° + cos 30°
4
=
1+ 3
4 3 −1
= × (to rationalise the denomiator)
1+ 3 3 −1
=2 ( 3 −1 )
= 1.46 (3 s.f.)
b R ↑ ()
T sin 60° − T sin 30° − mg = 0
mg = T (sin 60° − sin 30°)
3 1
=2 ( )
3 − 1
2
− (using T = 2
2
( )
3 − 1 from a)
( )
2
= 3 −1
= 4−2 3
m=
(4 − 2 3)
g
= 0.055 kg = 55g
12 12 5
7 tanθ = ⇒ sin θ = and cos θ =
5 13 13
Let the normal reaction be R N.
a R(→)
P cos θ − 1 = 0
1
∴P =
cos θ
13
=
5
P = 2.6
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 4
7 b R ↑ ()
R − P sin θ − 2 = 0
∴ R = P sin θ + 2
12
= 2.6 × + 2
13
= 2.4 + 2
= 4.4
8 a Consider the particle of mass 2m kg first, as it has only two forces
acting on it. This enables you to find the tension.
()
R ↑
T − 2mg = 0
∴T = 2mg
Consider the particle of mass m kg:
R(→)
T −F =0
∴ F = T = 2mg
= 19.6m
()
R ↑
R − mg = 0
∴ R = mg
= 9.8m
b Let T ' be the new tension in the string.
Consider the particle of mas 2m kg:
()
R ↑ : T ' = 2mg
Consider the particle of mass m kg:
R(→)
T 'cos 30° − F ' = 0
3
∴ F ' = 2mg ×
2
= 3mg
= 17m (2 s.f.)
()
R ↑
R '+ T 'sin 30 − mg = 0
∴ R ' = mg − T 'sin 30
= mg − 2mg × 1 (using T ' = 2mg )
2
=0
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 5
9 Let the normal reaction be R N.
R (ր) :
P − 2 g sin 45° = 0
∴ P = 2 g sin 45°
=g 2
= 14 N (2 s.f.)
10 Let the normal reaction be R N.
R (ր) :
P cos 45° − 4 g sin 45° = 0
4 g sin 45°
∴P =
cos 45°
= 4g
= 39 (2 s.f.)
11 a Let the normal reaction between the particle P and the plane be R N.
Let the tension in the string be T N.
Consider first the 5 kg mass.
()
R ↑
T − 5g = 0
∴T = 5g
Consider the 2 kg mass.
R ( տ)
R − 2 g cos θ = 0
4
R = 2g ×
5
8g
=
5
= 16 N (2 s.f.)
b R (ր)
T − F − 2 g sin θ = 0
F = T − 2 g sin θ
= 5 g − 2 g × 3 (using T = 5 g from above)
5
19 g
=
5
= 37 N (2 s.f.)
c Assuming the pulley is smooth means there is no friction between it and the string.
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 6
12 Let the normal reaction be R N.
First, resolve along the plane to find P as it is the only unknown when
resolving in that direction.
R (ր)
P cos 30° − 5cos 45° − 20sin 45° = 0
5cos 45° + 20sin 45°
∴P =
cos 30°
2 2 2
= 5 × + 20 × ×
2 2 3
25 2
=
3
25 6
=
3
= 20.4 (3 s.f.)
R ( տ)
R + P sin 30° + 5sin 45° − 20 cos 45° = 0
25 6
R = 20 cos 45° − 5sin 45° − P sin 30° (as P = )
3
15 25 6
R= −
2 6
45 2 − 25 6
=
6
= 0.400 (3 s.f.)
© Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free. 7
