Answer:
m = 1.154
P = -0.923
Step-by-step explanation:
A particle of mass m kg rests on a smooth slope that is inclined at an angle α to the horizontal. Two external forces act on the particle: a force of magnitude P N acts parallel to the horizontal (←), and another force of magnitude 12 N acts perpendicular to the horizontal (↑). The particle is accelerating at 1 m/s² up the plane.
Given that tan α = 0.75, then sin α = 0.6 and cos α = 0.8.
To find the values of m and P, begin by resolving forces parallel and perpendicular to the plane. (We resolve forces parallel and perpendicular to the plane, rather than horizontally and vertically because the object is on an inclined plane).
Resolve parallel to the plane (↗):
[tex]F_{\text{net}}=ma\\\\12\sin \alpha - P\cos \alpha - m\text{g}\sin \alpha = m \times 1\\\\12\times 0.6 - P \times 0.8 - m \times 9.8 \times 0.6 = m\\\\7.2 - 0.8P - 5.88m = m\\\\6.88m +0.8P = 7.2[/tex]
Weight always acts vertically downwards.
Resolve perpendicular to the plane (↖):
[tex]F_{\text{net}}=ma\\\\12\cos \alpha + P\sin \alpha - m\text{g}\cos \alpha = m \times 0\\\\12 \times 0.8 + P \times 0.6 - m \times 9.8 \times 0.8 = 0\\\\9.6 + 0.6P - 7.84m = 0\\\\7.84m - 0.6P = 9.6[/tex]
Now we have two equations involving m and p that can be solved simultaneously.
Rewrite the second equation to isolate P:
[tex]P=\dfrac{7.84m - 9.6}{0.6}[/tex]
Substitute this into the first equation and solve for m:
[tex]6.88m + 0.8\times \dfrac{7.84m - 9.6}{0.6} = 7.2\\\\\\6.88m + \dfrac{6.272m - 7.68}{0.6} = 7.2\\\\\\\dfrac{4.128m}{0.6} + \dfrac{6.272m - 7.68}{0.6} = 7.2\\\\\\\dfrac{10.4m - 7.68}{0.6} = 7.2\\\\\\10.4m-7.68=4.32\\\\\\10.4m=12\\\\\\m=\dfrac{15}{13}\\\\\\m=1.154\; \sf (3 \;d.p.)[/tex]
Substitute the exact value of m into the expression for P and solve for P:
[tex]P=\dfrac{7.84\left(\dfrac{15}{13}\right) - 9.6}{0.6}\\\\\\P=\dfrac{\dfrac{588}{65} - 9.6}{0.6}\\\\\\P=-\dfrac{180}{195}\\\\\\P=-\dfrac{12}{13}\\\\\\P=-0.923\; \sf (3\;d.p.)[/tex]
