Physics- Vectors topic
1. A body of mass 2kg is suspended from a rigid support. The body is held at 60° with the rest position by applying force F. If T is the tension in the string F/T is
a. 1
b. 2/√3
c. √3/2
d. 1/√3
@. A body of weight √3 is suspended vertically using a rope and is pulled horizontally such that the rope makes an angle 30° to the vertical Then the tension that rope is
a. 3N
b. 2N
c. √3N
d. 2√3N
3. 11 force each of 5N acts on a particle simultaneously , if each force makes an angle 30° with the next one, the resultant of all forces is
a. 55N
b. 11N
c. 15N
d. 5N
4. A body of 10kg is suspended by a rope and pulled to a side by a force so that the rope makes an angle 60° with vertical. Find the horizontal force( taking g = 10m/s²)
a. 100√3N
b. 100N
c. 50√3N
d. 200N
5. A body of mass 20kg is suspended vertically by means of a string of length 10cm , the horizontal force required to pull the mass horizontally through 6 cm is
a. 100N
b. 150N
c. 200N
d. 250N
Irrelevant answers will be reported.
Answers
Answer :
1) option (c) √3/2
2) option (b) 2 N
3) option (d) = 5 N
4) option (a) = 100√3 N
5) 120 N
Explanation :
(refer to the attachment)
Let
T be the tension in the string
F is the applied horizontal force
θ is the angle made by the string with the vertical
m is the mass of the body suspended
L is the length of the string
x is the displaced length
Question - 1 :
- Given,
m = 2 kg
θ = 60°
- To find,
F/T = ?
- Solution,
From the attachment,
F/T = sin 60°
F/T = √3/2
Question - 2 :
- Given,
Weight, mg = √3 N
θ = 30°
- To find.
Tension in the rope, T = ?
- Solution,
From the attachment,
√3 = T cos 30°
√3 = T (√3/2)
1 = T/2
T = 2 N
Question - 3 :
- Given,
no. of forces acting on the body = 11
each force = 5 N
angle made by a force with the next one = 30°
- To find,
resultant force = ?
- Solution,
For the resultant to be zero, no.of minimum forces required is
But we're given, no.of forces acting on the body as 11
The polygon we get is having 1 side (12 − 11 = 1) lesser than the polygon with resultant zero.
Hence, the resultant is 5 N
Question - 4 :
- Given,
m = 10 kg
θ = 60°
g = 10 m/s²
- To find,
horizontal force, F = ?
- Solution,
Refer to the attachment,
tan 60° = F/(10×10)
√3 = F/100
F = 100 × √3
F = 100√3 N
Question - 5 :
- Given,
m = 20 kg
L = 10 cm
x = 6 cm
- To find,
horizontal force, F = ?
- Solution,
Refer to the attachment,
x/L = F/mg
6/10 = F/(20×10)
3/5 = F/200
F = (3/5) × 200
F = 3 × 40
F = 120 N
Answer:
1) option (c) 3/2
2) option (b) 2 N
3) option (d) = 5 N
4) option (a) = 100v3 N
5) 120 N
Explanation :
(refer to the attachment)
Let
T be the tension in the string
F is the applied horizontal force
e is the angle made by the string with the vertical
m is the mass of the body suspended
L is the length of the string
x is the displaced length
Question - 1:
Given,
m = 2 kg
e = 60°
• To find,
F/T = ?
. Solution,
From the attachment,
F = T sin 0
F/T = sin 60°
F/T = 3/2
Question - 2:
Given,
Weight, mg = 3 N
e = 30°
• To find.
Tension in the rope, T = ?
• Solution,
From the attachment,
mg = T cos 0
V3 = T cos 30
V3 = T (V3/2)
1 = T/2
T= 2 N
Question - 3:
Given,
no. of forces acting on the body = 11
each force = 5 N
angle made by a force with the next one = 30°
• To find,
resultant force = ?
Solution,
For the resultant to be zero, no.of minimum forces required
is
n =
360°
30°
= 12
But we're given, no.of forces acting on the body as 11
The polygon we get is having 1 side (12 11 = 1) lesser than the polygon with resultant zero.
Hence, the resultant is 5 N
Question - 4:
Given,
m = 10 kg
= 60°
g = 10 m/s2
• To find,
horizontal force, F = ?
Solution,
Refer to the attachment,
F
mg
tan 60° = F/(10×10)
v3 = F/100
F = 100 x v3
F = 100v3 N
Question - 5:
Given,
m= 20 kg
L = 10 cm
X= 6 cm
To find
m = 20 kg
L = 10 cm
x= 6 cm
• To find,
horizontal force, F = ?
• Solution,
Refer to the attachment,
X
L
F
mg
tane =
x/L = F/mg
6/10 = F/(2010)
3/5 = F/200
F = (3/5) x 200
F= 3 x 40
F= 120 N
