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
Answer:
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Explanation:
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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
=μ
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
=μ
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.
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