Author Topic: ehh (Read 1307 times) Tweet Share

fredrick · « **on:** August 16, 2008, 07:44:10 pm »

Questions i had trouble with

First Q)
In a lift accelerating downwards at 1ms^-2, a spring balance shows apparent weight of a body to be 2.5 kg wt. What would be he reading if the lift was:
a)at rest and b) accelerating up at 2ms^-2.

Next Q)
A body is projected up an incline of 20 degrees with a velocity of 10m/s. If the coefficint of friction between the body and the plane is 0.25, find the distance it goes up the plane and the velocity with which it returns to its starting point.

Next)
A body of mass 5 kg is placved on a smooth horizontal plane and is acted upon by the following 3 horizontal forces:
1)a force of 8N in direction 330 degrees
2)a force of 10N in direction 090 degrees
3)a frce of PN in direction 180 degrees
Given that the magnitude of acceleration of the body is 2ms^-2, calculate the value of P.

Thats all, for now. Cheers!

Mao · « **Reply #1 on:** August 16, 2008, 08:19:47 pm »

1)
apparent weight is a force [and in this case a Normal reaction force]

hence, 2.5 kg wt = 2.5g N

$2.5g=m\cdot \left(g-1ms^-2\right)\implies m\approx \frac{2.5\times 9.8}{8.8}\approx 2.784kg$

at rest: weight is $W=mg=\frac{2.5g^2}{g-1}\approx 2.784\cdot 9.8\approx 27.3\; kg\; wt$

accelerating upwards at 2ms^-2: weight is $W=mg=\frac{2.5g(g+2)}{g-1}\approx 2.784 \cdot (9.8+2)\approx 32.85\; kg\; wt$

2) the forces acting are: gravitational force $F_g=mg\sin \theta$ , normal reaction force (no impact on motion) $F_N=mg\cos \theta$ , and the friction force $F_f=\mu F_N$ , where $\mu$ is the coefficient of friction

net acceleration going up the incline: $a_{up}=-(F_g+F_f)=-g\left( \sin 20^o+\frac{\cos 20^o}{4}\right)$

using SUVAT: $v^2-u^2=2as$

$\implies 100=g\left( \sin 20^o+\frac{\cos 20^o}{4}\right)\cdot s$

$\implies s\approx 17.69m$ (i hope)

now accelerating down: $a_{down}=F_g-F_f=g\left( \sin 20^o-\frac{\cos 20^o}{4}\right)$

using SUVA: $v^2-u^2=2as$

$\implies v^2=2\cdot g\left( \sin 20^o-\frac{\cos 20^o}{4}\right)\cdot s$

$\implies v\approx 6.09ms^{-1}$ (i hope)

3)
resolving into components [0 degrees and 90 degrees]

8N: 0 degrees: $4\sqrt{3}N$ , 90 degrees: -4N
10N: 0 degrees: 0N, 90 degrees: 10N
PN: 0 degrees: -PN, 90 degrees: 0N

resultant at 0 degrees: $4\sqrt{3}-P$ N
resultant at 90 degrees: 6N

magintude of resultant force is 10N:

$(4\sqrt{3}-P)^2+36=100$

$4\sqrt{3}-P=\pm 8$

$P=4\sqrt{3}\pm 8$ N

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fredrick

ehh

Mao

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