A body of mass 5kg is acted upon by two perpendicular force of 4N and 3N. Find the magnitude of acceleration.
Answers
Step-by-step explanation:
Step 1: Drawing FBD [Refer Fig. 1]
Step 2: Finding Resultant Force [Refer Fig. 2]
Magnitude of Resultant:
Since F
1
and F
2
are perpendicular
∴ Magnitude of Resultant force on body is given by
R=
F
1
2
+F
2
2
=
8
2
+6
2
N =10N
Direction of Resultant:
θ =tan
−1
F
1
F
2
[Refer Fig. 2]
=tan
−1
8
6
=36.9
o
≈37
o
Step 3: Applying Newton’s second Law for finding acceleration [Refer Fig. 3]
F
net
=ma
⇒ 10=5a
⇒ ∣
a
∣=2m/s
2
Hence magnitudes of acceleration is 2m/s
2
directed at 37
o
from the 8 N force
Answer:
Step 1: Drawing FBD [Refer Fig. 1]
Step 2: Finding Resultant Force [Refer Fig. 2]
Magnitude of Resultant:
Since F
1
and F
2
are perpendicular
∴ Magnitude of Resultant force on body is given by
R=
F
1
2
+F
2
2
=
8
2
+6
2
N =10N
Direction of Resultant:
θ =tan
−1
F
1
F
2
[Refer Fig. 2]
=tan
−1
8
6
=36.9
o
≈37
o
Step 3: Applying Newton’s second Law for finding acceleration [Refer Fig. 3]
F
net
=ma
⇒ 10=5a
⇒ ∣
a
∣=2m/s
2
Hence magnitudes of acceleration is 2m/s
2
directed at 37
o
from the 8 N force