Resolving Forces

Original post by MWorldwide19

https://www.chegg.com/homework-help/questions-and-answers/6-sledge-mass-10-mathrm-~kg-pulled-along-rough-horizontal-plane-force-p-acts-angle-45-circ-q123954094

Just curious if the answer to the question I've linked is 12.6 (3sf) as I don't want to pay for the answer. Can't find the answer anywhere else so hope someone here can help me if I haven't got it right. Thanks!

getting a bit higher, can you upload what you did?

Reply 2

OP

Original post by mqb2766

Original post by MWorldwide19

https://www.chegg.com/homework-help/questions-and-answers/6-sledge-mass-10-mathrm-~kg-pulled-along-rough-horizontal-plane-force-p-acts-angle-45-circ-q123954094

Just curious if the answer to the question I've linked is 12.6 (3sf) as I don't want to pay for the answer. Can't find the answer anywhere else so hope someone here can help me if I haven't got it right. Thanks!

getting a bit higher, can you upload what you did?

Sure but I don't think I can upload an image so I'll just talk through it

I first wrote out R+Psin45= 10g which meant that R= 10g-Psin45 or 98-(2^½/2)P
I then solved for Fmax= uR which gave Pcos45= 0.1 × 10g-Psin45
This led to me getting (2^½/2)P= 9.8-(2^½/20)P when simplifying
Finally I solved this equation for P which got me 12.6

Reply 3

Original post by MWorldwide19

Sure but I don't think I can upload an image so I'll just talk through it

I first wrote out R+Psin45= 10g which meant that R= 10g-Psin45 or 98-(2^½/2)P
I then solved for Fmax= uR which gave Pcos45= 0.1 × 10g-Psin45
This led to me getting (2^½/2)P= 9.8-(2^½/20)P when simplifying
Finally I solved this equation for P which got me 12.6

Youve not accounted for the fact its accelerating at 0.3 m/s^2 when resolving horizontally

(edited 4 months ago)

Reply 4

OP

Original post by mqb2766

Original post by MWorldwide19

Sure but I don't think I can upload an image so I'll just talk through it

I first wrote out R+Psin45= 10g which meant that R= 10g-Psin45 or 98-(2^½/2)P
I then solved for Fmax= uR which gave Pcos45= 0.1 × 10g-Psin45
This led to me getting (2^½/2)P= 9.8-(2^½/20)P when simplifying
Finally I solved this equation for P which got me 12.6

Youve not accounted for the fact its accelerating at 0.3 m/s^2 when resolving horizontally

Oh wait so does that mean I'd have to include it when I do Fr=Pcos45? So it would be Pcos45+0.3=Fr?

Reply 5

Original post by MWorldwide19

Oh wait so does that mean I'd have to include it when I do Fr=Pcos45? So it would be Pcos45+0.3=Fr?

Its simply newton 2 so horizontally
ma = net force
Youre trying to add an acceleration 0.3 to forces which cant be right. So the net force is obv
Pcos(45) - muR

A sketch of the forces, both vertically and horizontally, would really help to get directions / signs / acceleration clear. Intuitively, there must be a 3N excess force to accelerate a 10kg object at 0.3 m/s^2.

(edited 4 months ago)

Reply 6

OP

Original post by mqb2766

Original post by MWorldwide19

Oh wait so does that mean I'd have to include it when I do Fr=Pcos45? So it would be Pcos45+0.3=Fr?

Its simply newton 2 so horizontally
ma = net force
Youre trying to add an acceleration 0.3 to forces which cant be right. So the net force is obv
Pcos(45) - muR

A sketch of the forces, both vertically and horizontally, would really help to get directions / signs / acceleration clear.

Oh okay I get it now so is the answer instead 16.5 (3sf)?

Reply 7

Original post by MWorldwide19

Oh okay I get it now so is the answer instead 16.5 (3sf)?

I think thats what I got.

(edited 4 months ago)

Reply 8

OP

Original post by mqb2766

Original post by MWorldwide19

Oh okay I get it now so is the answer instead 16.5 (3sf)?

I think thats what I got.

Okay great, thanks for your help!

Reply 9