M1 q

PLease can anyone help.

1. A particle of mass m lies on a smooth plane inclined at angle A to the horizontal. It is held in equilibrium by a string, which lies in a vertical plane through a line of greatest slope and makes an angle of B with the plane. The tension in string is of magnitude T, and force exerted by the plane on the particle is of magnitude R.

a) Find in terms of m, g, A and B, an expression for T.
b) Show that R = mg cosA (1 - tan A tan B)

Reply 1

habosh

13

confused?

PLease can anyone help.

1. A particle of mass m lies on a smooth plane inclined at angle A to the horizontal. It is held in equilibrium by a string, which lies in a vertical plane through a line of greatest slope and makes an angle of B with the plane. The tension in string is of magnitude T, and force exerted by the plane on the particle is of magnitude R.

a) Find in terms of m, g, A and B, an expression for T.
b) Show that R = mg cosA (1 - tan A tan B)

I'm not sure a bout (b)so I'll try to solve a
by dorwing a diagram
you can see that
the angle the weight on the particle makes with the vertical equals A
so
T=mgsinA/cosB

I'll try B
as we see the normal eqauls mgcosA,but I dunno how to obtain (1-tanAtanB)

Reply 2

Jonny W

8

Draw a diagram!

--

Resolving parallel to the slope,

T cos(B) = mg sin(A)

So the answer to (a) is T = mg sin(A)/cos(B).

--

Resolving perpendicularly to the slope,

R + T sin(B) = mg cos(A)

So

R
= mg cos(A) - T sin(B)
= mg cos(A) - mg sin(A)sin(B)/cos(B)
= mg cos(A) [1 - sin(A)sin(B)/(cos(A)cos(B))]
= mg cos(A) [1 - tan(A)tan(B)]

Reply 3

habosh

13

Jonny W

Draw a diagram!

--

Resolving parallel to the slope,

T cos(B) = mg sin(A)

So the answer to (a) is T = mg sin(A)/cos(B).

--

Resolving perpendicularly to the slope,

R + T sin(B) = mg cos(A)

So

R
= mg cos(A) - T sin(B)
= mg cos(A) - mg sin(A)sin(B)/cos(B)
= mg cos(A) [1 - sin(A)sin(B)/(cos(A)cos(B))]
= mg cos(A) [1 - tan(A)tan(B)]