A 1.50 kg block slides up a ramp that is inclined at 30 degrees above the horizontal. A constant horizontal force F is applied to the block as shown. Resolve all the forces intox-and-y components. 1 resolve all the forces into x-and y- components and draw the components on the diagram below can you guys help me doing this. A 1.50 kg block slides up a ramp that is inclined at 30.0° above the horizontal. Aconstant horizontal force F is applied to the block as shown.F-30The free-body diagram for this situation is shown below:F stands for the horizontal force. It acts parallel to the floor.n stands for the normal force. It acts perpendicular to the ramp.f stands for the friction force. It acts parallel to the incline.mg is the weight vector. It acts perpendicular to the floor.(1) Resolve all the forces into x- and y-components and draw the components onthe diagram below.+ynFF30"mgmg(2) In this box, write an equation for the sum of the force in the x- (3) In this box, write an equation for the sum ofthe forces in the direction. You should havean algebraic equation-variables only! Writeall components in terms of sine and cosinedirection. You should have an algebraic equation-variablesonly!! Write all components in terms of sine and cosinefunctions.functions.EEx =ma全 FY_=mamy+n - mg cos 300 sF + Fcos 30g sih 30"1)

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Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter5: Newton's Laws Of Motion

Section5.7: Some Specific Forces

Problem 5.9CE: For all three situations, find the magnitude and direction of the tension force(s) exerted on...

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A 1.50 kg block slides up a ramp that is inclined at 30 degrees above the horizontal. A constant horizontal force F is applied to the block as shown. Resolve all the forces intox-and-y components. 1 resolve all the forces into x-and y- components and draw the components on the diagram below can you guys help me doing this.

A 1.50 kg block slides up a ramp that is inclined at 30.0° above the horizontal. A
constant horizontal force F is applied to the block as shown.
F-
30
The free-body diagram for this situation is shown below:
F stands for the horizontal force. It acts parallel to the floor.
n stands for the normal force. It acts perpendicular to the ramp.
f stands for the friction force. It acts parallel to the incline.
mg is the weight vector. It acts perpendicular to the floor.
(1) Resolve all the forces into x- and y-
components and draw the components on
the diagram below.
+y
n
F
F
30
"mg
mg
(2) In this box, write an equation for the sum of the force in the x- (3) In this box, write an equation for the sum of
the forces in the direction. You should have
an algebraic equation-variables only! Write
all components in terms of sine and cosine
direction. You should have an algebraic equation-variables
only!! Write all components in terms of sine and cosine
functions.
functions.
EEx =ma
全 FY_
=ma
my
+n - mg cos 300 s
F + Fcos 30
g sih 30"
1)

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