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Urgent* Two particals, find the velocity for each.

+3
306
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+1314

Urgent* Two particals, find the velocity for each.
Note: the v1 and v2 are velocities.

17= v1sin(30)+v2sin(-45)  Along the x axis.

Substituting in the y axis equation where v2 = v1cos(30)/cos(45)

v2sin(30) -v1cos(30)/cos(45) *cos45 = 17

Solve please.

Stu  Jul 18, 2014

Best Answer

#7
+26329
+10

I think you might be trying to solve the following equations (from a previous problem):

17 = v1*cos(45) + v2*cos(30)  ...(1)

0 = -v1*sin(45) + v2*sin(30)     ...(2)

Add v1*sin(45) to both sides of (2) and then divide both sides by sin(45)

v1 = v2*sin(30)/sin(45)             ...(3)

Replace v1 in (1) by (3)

17 =  v2*sin(30)/sin(45)*cos(45) + v2*cos(30)

Factor the right-hand side

17 =  v2*[sin(30)/sin(45)*cos(45) + cos(30)]

Divide both sides by [sin(30)/sin(45)*cos(45) + cos(30)] to get v2

v2 = 17/[sin(30)/sin(45)*cos(45) + cos(30)]   ...(4)

Replace v2 in (3) by (4) to get v1

v1 = 17/[sin(30)/sin(45)*cos(45) + cos(30)]*sin(30)/sin(45)

$${\mathtt{v1}} = {\frac{{\frac{{\mathtt{17}}}{\left[{\frac{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}\right]}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}}} \Rightarrow {\mathtt{v1}} = {\mathtt{8.799\: \!847\: \!533\: \!482\: \!900\: \!7}}$$

$${\mathtt{v2}} = {\frac{{\mathtt{17}}}{\left[{\frac{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}\right]}} \Rightarrow {\mathtt{v2}} = {\mathtt{12.444\: \!863\: \!728\: \!674\: \!910\: \!2}}$$

Alan  Jul 18, 2014
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2+0 Answers

#7
+26329
+10
Best Answer

I think you might be trying to solve the following equations (from a previous problem):

17 = v1*cos(45) + v2*cos(30)  ...(1)

0 = -v1*sin(45) + v2*sin(30)     ...(2)

Add v1*sin(45) to both sides of (2) and then divide both sides by sin(45)

v1 = v2*sin(30)/sin(45)             ...(3)

Replace v1 in (1) by (3)

17 =  v2*sin(30)/sin(45)*cos(45) + v2*cos(30)

Factor the right-hand side

17 =  v2*[sin(30)/sin(45)*cos(45) + cos(30)]

Divide both sides by [sin(30)/sin(45)*cos(45) + cos(30)] to get v2

v2 = 17/[sin(30)/sin(45)*cos(45) + cos(30)]   ...(4)

Replace v2 in (3) by (4) to get v1

v1 = 17/[sin(30)/sin(45)*cos(45) + cos(30)]*sin(30)/sin(45)

$${\mathtt{v1}} = {\frac{{\frac{{\mathtt{17}}}{\left[{\frac{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}\right]}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}}} \Rightarrow {\mathtt{v1}} = {\mathtt{8.799\: \!847\: \!533\: \!482\: \!900\: \!7}}$$

$${\mathtt{v2}} = {\frac{{\mathtt{17}}}{\left[{\frac{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}\right]}} \Rightarrow {\mathtt{v2}} = {\mathtt{12.444\: \!863\: \!728\: \!674\: \!910\: \!2}}$$

Alan  Jul 18, 2014
#8
+1314
0

Thanks. Now I shouldnt get lost.

Stu  Jul 21, 2014

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