Physics, asked by chandramanikumari098, 13 hours ago

g. Block resting on on the o able is attached to a & 150N smooth chorizonytal plane shown in figine it a horizontal force force 'pi block is applied detamine! i the tension in the eable and reaction everted b by (P=15 N )​

Answers

Answered by arthanarivishnu2008
0

Answer:

VG GG fvdcjsclbwfkbcsbieib

evmsvnsvnisibsu

Explanation:

ehsvkebkeibecniecnineciecbiecbi

Answered by tabassumsultana1978
0

Answer:

conditions of equilibrium i.e. \sum H = 0∑H=0 along the inclined plane, and \sum V = 0∑V=0 perpendicular to the inclined plane. Free body diagram of the block is shown in figure Ex 7(a).

\sum H = 0: P \cos 45° + F – 150 \sin 45° = 0 ∑H=0:Pcos45°+F–150sin45°=0

Now for impending motion from (Eq. 5.1), we know F_{max} = \mu_{s} N.F

max

s

N.

\therefore P \cos 45° + \mu_{s} N – 150 \sin 45° = 0∴ Pcos45°+μ

s

N–150sin45°=0 …(Eq. 1)

Now, \sum V = 0 : N – 150 \cos 45° – P \sin 45° = 0∑V=0:N–150cos45°–Psin45°=0

\therefore N = 150 \cos 45° + P \sin 45°∴ N=150cos45°+Psin45°

Substituting value of N in to Eq. 1,

P \cos 45° + 0.25 (150 \cos 45° + P \sin 45°) – 150 \sin 45° = 0Pcos45°+0.25(150cos45°+Psin45°)–150sin45°=0

\therefore P = P_{min} = 90 N∴ P=P

min

=90N

Now consider impending motion of the block up the plane and then apply conditions of equilibrium i.e. \sum H = 0∑H=0 along the inclined plane, and \sum V = 0∑V=0 perpendicular to the inclined plane. Free body diagram of the block is shown infigure Ex 7(b).

\sum H = 0: P \cos 45° – F – 150 \sin 45° = 0 ∑H=0:Pcos45°–F–150sin45°=0

Now for impending motion from (Eq. 5.1), we know F_{max} = \mu_{s} NF

max

s

N.

\therefore P \cos 45° – \mu_{s} N – 150 \sin 45° = 0∴ Pcos45°–μ

s

N–150sin45°=0 …(Eq. 2)

Now, \sum V = 0 : N – 150 \cos 45° – P \sin 45° = 0∑V=0:N–150cos45°–Psin45°=0

\therefore N = 150 \cos 45° + P \sin 45°∴ N=150cos45°+Psin45°

Substituting value of N in to Eq. 2,

P \cos 45° – 0.25 (150 \cos 45° + P \sin 45°) – 150 \sin 45° = 0Pcos45°–0.25(150cos45°+Psin45°)–150sin45°=0

\therefore P = P_{max} = 250 N ∴ P=P

max

=250N

\therefore ∴ Range of horizontal force P is from 90 N to 250 N that can keep the block in equilibrium.

mark me brainlist pls

Similar questions