## Homework explain

**Two pressures act on a 55 kg object. One has actually a size of 65 N command 59 levels clockwise indigenous the pos. X axis. The other has magnitude 35 N at 32 o clockwise indigenous the pos. Y axis.What is the acceleration the the object?The prize is given as 1.1 m/ s squared,(I gained a similar answer 1.0 m/s squared.) , however not confident that ns approached the equipment correctly.Thanks in advance for any aid :-)2. Homework Equations**

SDTK, you wrote:ΣFx = sin(32)(35) + cos(59)(65) = max620.19 = maxHow walk you get from the an initial line come the second line?Edit: it looks prefer you may have actually multiplied those 2 terms, fairly than include them.

You are watching: Two forces act on a 55 kg object

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The difficulty states the two pressures act top top the object. Heaviness was not one of them.Cartesian vector components include in quadrature: square root of the sum of the squares.

SDTK, friend wrote:ΣFx = sin(32)(35) + cos(59)(65) = max620.19 = maxHow walk you gain from the first line come the second line?Edit: it looks prefer you may have actually multiplied those 2 terms, rather than include them.

The difficulty states the two forces act on the object. Gravity was not one of them.Cartesian vector components include in quadrature: square root of the amount of the squares.

gneillThank you! Is this what you are telling me? (work presented below)(I"m concerned due to the fact that I get 0.99 m/s^2, i m sorry is different from the answer of

Find x component of the network force, then uncover it in Y direction. THESE room of course 90 levels wrt each other. Calculate the magnitude of the complete net force. Then calculate a indigenous F and m.

No. The given forces won"t in general kind a best angle triangle (and lock don"t in this case). You should sum your components.What i was acquiring at was, once you"ve summed the individual x and y contents of the vectors to type the resultant vector"s components, climate the magnitude of the resultant is found by summing that is x and also y components. Friend still require to uncover the components of the result vector utilizing the an approach that you used in article #6 (only forget the gravitational force since it theatre no role here).

No. The given pressures won"t in general kind a best angle triangle (and they don"t in this case). You need to sum your components.What ns was obtaining at was, as soon as you"ve synthetic the separation, personal, instance x and y materials of the vectors to form the result vector"s components, climate the magnitude of the resultant is discovered by summing the x and also y components. Friend still need to discover the materials of the result vector using the an approach that you used in post #6 (only forget the gravitational force because it dram no role here).

gneillThank you!Is this what you are telling me? (work displayed below)(I"m concerned due to the fact that I get 0.99 m/s^2, i m sorry is various from the answers of

ObjectivelyRational 1.0577m/s^2, and the answer provided with the trouble 1.1m/s^2.)ThanksView attachment 108525

In the triangle of forces you drew, what is the angle between the 65N and the 35N? just a little of elementary school geometry needed.

Yes, if you have actually the angle in between the forces. In basic though I suppose a college student to be confronted with summing multiple pressures to find a resultant. Doing lock one at a time via the cosine rule can be tedious.

Thank you!I uncovered f net x to be 67.92N sin(32)35N + cos(59)65Nf network y to be -26.03 N cos(32)35N - sin(59)65NIf I attract a vector, size of 68N ~ above the positive x axis, and also a vector magnitude 26N ~ above the an unfavorable y axis,... Then attract a vector in between the two that meets the segment representing the vector sum, is that a vector representing the total net force?

Find x component of the network force, then uncover it in Y direction. THESE space of food 90 degrees wrt every other. Calculation the size of the total net force. Then calculate a indigenous F and m.

Find x component of the net force, then discover it in Y direction. THESE room of food 90 degrees wrt each other. Calculation the magnitude of the full net force. Then calculate a native F and also m.

Find x ingredient of the net force, then uncover it in Y direction. THESE space of food 90 levels wrt each other. Calculate the magnitude of the total net force. Then calculation a from F and m.

In the triangle of forces you drew, what is the angle in between the 65N and the 35N? just a little bit of elementary geometry needed.

Cartesian materials of a vector type a rectangle through the political parties paralleling the name: coordinates axes and also the vector itself developing the diagonal:

Note the the contents ##f_x## and ##f_y## that vector ##f## space at appropriate angles to each other and are parallel come the axes. The magnitude of f is ##|f| = \sqrtf_x^2 + f_y^2##I don"t agree v the value of the x_component that you calculate in your short article #17. Deserve to you give much more details of the calculation? The y_component looks okay.

Ok, for this reason you have a selection of approaches. For a simple resultant of two vectors, you can use the angle between them (when drawn in the nose-to-tail format of the pressure triangle) and also apply the cosine rule, or for multiple used forces you can reduce every to the X and also Y components and also find the amount along every axis.

thank you, give thanks to you, give thanks to you! i recalculated the F(net) x, acquiring 52 N.The illustration you common helped! :-)

Ok, for this reason you have actually a an option of approaches. Because that a an easy resultant of 2 vectors, you can use the angle between them (when attracted in the nose-to-tail format of the pressure triangle) and apply the cosine rule, or because that multiple used forces you deserve to reduce each to that X and Y components and also find the amount along each axis.

See more: Height Of A Standard Coffee Cup Size? How Tall Is A 12 Oz Mug

thank you! i did not understand the cosine rule. I looked it up, and see how works, and it offers me the best answer! :-)

Cartesian materials of a vector form a rectangle through the sides paralleling the coordinate axes and also the vector itself forming the diagonal:View attachments 108530Note the the components ##f_x## and ##f_y## the vector ##f## room at best angles to every other and are parallel come the axes.The size of f is ##|f| = \sqrtf_x^2 + f_y^2##I don"t agree with the worth of the x_component that you calculated in your short article #17. Have the right to you give more details of that calculation? The y_component watch okay.

gneil,thank you! i posted a reply, however it looks choose it went together a separate posting instead of a direct reply. I view my error in the x_ ingredient calculation. Ns (now) have 52N for F(net)x. Give thanks to you additionally for the notes around the Cartesian contents for the net pressure vector!

gneil,thank you! i posted a reply, yet it looks favor it went as a different posting rather of a straight reply. I see my error in the x_ ingredient calculation. I (now) have actually 52N for F(net)x. Give thanks to you also for the notes around the Cartesian contents for the net force vector!

## The attempt at a Solution

-- Is the correct strategy to find the sum of: every x components, and y materials of the forces, (65N, 35N, and weight)?SDTK, you wrote:ΣFx = sin(32)(35) + cos(59)(65) = max620.19 = maxHow walk you get from the an initial line come the second line?Edit: it looks prefer you may have actually multiplied those 2 terms, fairly than include them.

You are watching: Two forces act on a 55 kg object

LikesSDTK

The difficulty states the two pressures act top top the object. Heaviness was not one of them.Cartesian vector components include in quadrature: square root of the sum of the squares.

SDTK, friend wrote:ΣFx = sin(32)(35) + cos(59)(65) = max620.19 = maxHow walk you gain from the first line come the second line?Edit: it looks prefer you may have actually multiplied those 2 terms, rather than include them.

The difficulty states the two forces act on the object. Gravity was not one of them.Cartesian vector components include in quadrature: square root of the amount of the squares.

gneillThank you! Is this what you are telling me? (work presented below)(I"m concerned due to the fact that I get 0.99 m/s^2, i m sorry is different from the answer of

Find x component of the network force, then uncover it in Y direction. THESE room of course 90 levels wrt each other. Calculate the magnitude of the complete net force. Then calculate a indigenous F and m.

No. The given forces won"t in general kind a best angle triangle (and lock don"t in this case). You should sum your components.What i was acquiring at was, once you"ve summed the individual x and y contents of the vectors to type the resultant vector"s components, climate the magnitude of the resultant is found by summing that is x and also y components. Friend still require to uncover the components of the result vector utilizing the an approach that you used in article #6 (only forget the gravitational force since it theatre no role here).

No. The given pressures won"t in general kind a best angle triangle (and they don"t in this case). You need to sum your components.What ns was obtaining at was, as soon as you"ve synthetic the separation, personal, instance x and y materials of the vectors to form the result vector"s components, climate the magnitude of the resultant is discovered by summing the x and also y components. Friend still need to discover the materials of the result vector using the an approach that you used in post #6 (only forget the gravitational force because it dram no role here).

gneillThank you!Is this what you are telling me? (work displayed below)(I"m concerned due to the fact that I get 0.99 m/s^2, i m sorry is various from the answers of

ObjectivelyRational 1.0577m/s^2, and the answer provided with the trouble 1.1m/s^2.)ThanksView attachment 108525

In the triangle of forces you drew, what is the angle between the 65N and the 35N? just a little of elementary school geometry needed.

Yes, if you have actually the angle in between the forces. In basic though I suppose a college student to be confronted with summing multiple pressures to find a resultant. Doing lock one at a time via the cosine rule can be tedious.

Thank you!I uncovered f net x to be 67.92N sin(32)35N + cos(59)65Nf network y to be -26.03 N cos(32)35N - sin(59)65NIf I attract a vector, size of 68N ~ above the positive x axis, and also a vector magnitude 26N ~ above the an unfavorable y axis,... Then attract a vector in between the two that meets the segment representing the vector sum, is that a vector representing the total net force?

Find x component of the network force, then uncover it in Y direction. THESE space of food 90 degrees wrt every other. Calculation the size of the total net force. Then calculate a indigenous F and m.

Find x component of the net force, then discover it in Y direction. THESE room of food 90 degrees wrt each other. Calculation the magnitude of the full net force. Then calculate a native F and also m.

Find x ingredient of the net force, then uncover it in Y direction. THESE space of food 90 levels wrt each other. Calculate the magnitude of the total net force. Then calculation a from F and m.

In the triangle of forces you drew, what is the angle in between the 65N and the 35N? just a little bit of elementary geometry needed.

Cartesian materials of a vector type a rectangle through the political parties paralleling the name: coordinates axes and also the vector itself developing the diagonal:

Note the the contents ##f_x## and ##f_y## that vector ##f## space at appropriate angles to each other and are parallel come the axes. The magnitude of f is ##|f| = \sqrtf_x^2 + f_y^2##I don"t agree v the value of the x_component that you calculate in your short article #17. Deserve to you give much more details of the calculation? The y_component looks okay.

Ok, for this reason you have a selection of approaches. For a simple resultant of two vectors, you can use the angle between them (when drawn in the nose-to-tail format of the pressure triangle) and also apply the cosine rule, or for multiple used forces you can reduce every to the X and also Y components and also find the amount along every axis.

thank you, give thanks to you, give thanks to you! i recalculated the F(net) x, acquiring 52 N.The illustration you common helped! :-)

Ok, for this reason you have actually a an option of approaches. Because that a an easy resultant of 2 vectors, you can use the angle between them (when attracted in the nose-to-tail format of the pressure triangle) and apply the cosine rule, or because that multiple used forces you deserve to reduce each to that X and Y components and also find the amount along each axis.

See more: Height Of A Standard Coffee Cup Size? How Tall Is A 12 Oz Mug

thank you! i did not understand the cosine rule. I looked it up, and see how works, and it offers me the best answer! :-)

Cartesian materials of a vector form a rectangle through the sides paralleling the coordinate axes and also the vector itself forming the diagonal:View attachments 108530Note the the components ##f_x## and ##f_y## the vector ##f## room at best angles to every other and are parallel come the axes.The size of f is ##|f| = \sqrtf_x^2 + f_y^2##I don"t agree with the worth of the x_component that you calculated in your short article #17. Have the right to you give more details of that calculation? The y_component watch okay.

gneil,thank you! i posted a reply, however it looks choose it went together a separate posting instead of a direct reply. I view my error in the x_ ingredient calculation. Ns (now) have 52N for F(net)x. Give thanks to you additionally for the notes around the Cartesian contents for the net pressure vector!

gneil,thank you! i posted a reply, yet it looks favor it went as a different posting rather of a straight reply. I see my error in the x_ ingredient calculation. I (now) have actually 52N for F(net)x. Give thanks to you also for the notes around the Cartesian contents for the net force vector!